Conservation of mechanical energy and the work–energy principle
Why is the first hill of a roller coaster always the highest of the track, even though the train is no longer pulled up after it?
Prior knowledge
- Mechanical energy is the sum of kinetic energy and potential energy, $E_\text{k} = \tfrac{1}{2} mv^2$ and $E_\text{p} = mgh$chapter Kinetic energy, potential energy and mechanical energy
- A force does work when an object moves in the direction of the force, and work transfers energychapter Work and energy
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 5.
Think about it
A roller coaster train leaves the station, and a chain lift slowly pulls it to the top of the first hill, at a height of $40\,\text{m}$. After that nothing pulls the train any more. The train plunges into the valley, speeds over the next hill and the one after that, and finally rolls to the brakes at the station. In the drawings of the track each hill is lower than the one before.
Where does the train get its speed without an engine, and why can no later hill be as high as the first?
Optional Review: kinetic energy
Predict
Make your predictions before investigating 3 questions
Prediction 1: empty or full car
Prediction 2: how much energy the resistive force takes
Optional Prediction 3: steep or curved track
Investigate
In the simulation a roller coaster car rolls through a valley. On the left is a snapshot with measuring posts, on the right the energies of the car as bars on the same scale.
Simulation
How to use
Choose the starting height $h_1$, the resistive force $F_\text{res}$, the mass of the car and the shape of the track. Press Start. The car starts from rest and moves until it stops for the first time. Then the readings show the speed in the middle of the bottom of the valley $v_\text{bottom}$ and the height $h_2$ the car rose to. Reset returns the car to the starting point. Speed slows down or speeds up the motion without changing the readings.
Assumptions of the model
The values are model values calculated by the simulation. The car is a point-like object that stays on the track. The bottom of the valley is $6\,\text{m}$ long and horizontal, and the slopes are either arcs with a radius of $21\,\text{m}$ or straight $45^\circ$ slopes. The joint between a straight slope and the bottom is assumed to be rounded so gently that the speed of the car does not change there. The resistive force $F_\text{res}$ stands for rolling resistance and air resistance together: it is constant in magnitude and always points against the motion. The normal force of the track is perpendicular to the motion, so it does no work.
How the calculation works
The motion of the car is calculated from Newton's second law along the track in small time steps. The energies are calculated from the state of the car: $E_\text{p} = mgh$ from the height $h$ above the bottom of the valley and $E_\text{k} = \tfrac{1}{2} mv^2$ from the speed, with $g = 9.81\,\text{m/s}^2$. $E_\text{mech} = E_\text{k} + E_\text{p}$, and $\Delta E_\text{mech}$ is its change from the starting moment. The work done by the resistive force is calculated from the definition $W_\text{other} = -F_\text{res}\,s$, where $s$ is the distance travelled along the track. In the bars $W_\text{other}$ is negative, so it is drawn downward from the zero level.
Experiment 1: mass of the car
Test the first prediction. The starting state chooses a starting height of $10\,\text{m}$ and a curved track without a resistive force. Change only the mass of the car. Press Start each time and record a measurement when the car has stopped.
Initial state: starting height $h_1$ = $10\,\text{m}$, resistive force $F_\text{res}$ = $0\,\text{N}$, track: curved, mass $m$: $400\,\text{kg}$. Kept constant: starting height $h_1$, resistive force $F_\text{res}$, track. Varied: mass $m$.
| mass $m$ | $v_\text{bottom}$ $(\text{m/s})$ | $h_2$ $(\text{m})$ |
|---|
Experiment 2: resistive force
Test the second prediction. The starting state chooses a starting height of $10\,\text{m}$, a mass of $400\,\text{kg}$ and a curved track. Change only the resistive force $F_\text{res}$, for example to the values $0$, $100$ and $200\,\text{N}$. Press Start each time and record a measurement when the car has stopped. For each row calculate the product $F_\text{res}\,s$ and compare it with the change in mechanical energy.
Initial state: starting height $h_1$ = $10\,\text{m}$, mass $m$: $400\,\text{kg}$, track: curved, resistive force $F_\text{res}$ = $0\,\text{N}$. Kept constant: starting height $h_1$, mass $m$, track. Varied: resistive force $F_\text{res}$.
| resistive force $F_\text{res}$ | $s$ $(\text{m})$ | $h_2$ $(\text{m})$ | $\Delta E_\text{mech}$ $(\text{J})$ |
|---|
Optional Experiment 3: shape of the track
Test the third prediction. The starting state chooses a starting height of $10\,\text{m}$, a mass of $400\,\text{kg}$ and a resistive force of $0\,\text{N}$. Change only the shape of the track. Press Start and record a measurement on both tracks.
Initial state: starting height $h_1$ = $10\,\text{m}$, mass $m$: $400\,\text{kg}$, resistive force $F_\text{res}$ = $0\,\text{N}$, track: curved. Kept constant: starting height $h_1$, mass $m$, resistive force $F_\text{res}$. Varied: track.
| track | $v_\text{bottom}$ $(\text{m/s})$ | $s$ $(\text{m})$ |
|---|
Conclusion
Use your measurements to explain when the mechanical energy of the car is conserved and by how much it decreases when a resistive force acts. Then explain why an empty and a full car are equally fast at the bottom of the valley when there is no resistive force.
Model answer
In experiment 1 there was no resistive force. With the values of the model the speed at the bottom was $14.0\,\text{m/s}$ for both the $400\,\text{kg}$ and the $800\,\text{kg}$ car, and the car rose back to a height of $10.0\,\text{m}$: the mechanical energy was conserved. The mass does not matter, because both sides of the equation $mgh = \tfrac{1}{2} mv^2$ are proportional to the mass, so $v = \sqrt{2gh}$. In experiment 2 a resistive force of $200\,\text{N}$ acted over a distance of $45.9\,\text{m}$, and the mechanical energy decreased by $9.19\,\text{kJ}$, which is exactly $200\,\text{N} \cdot 45.9\,\text{m}$. The car rose only to a height of $7.66\,\text{m}$. So the mechanical energy decreases by the work done by the resistive force, $\Delta E_\text{mech} = -F_\text{res}\,s$. The lost mechanical energy turned into internal energy of the wheels, the rails and the air.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Theory summary
Key idea
Mechanical energy changes its form
In the chapter Kinetic energy, potential energy and mechanical energy you learned that the mechanical energy of an object is the sum of its kinetic energy and potential energy, $E_\text{mech} = E_\text{k} + E_\text{p}$. When a roller coaster car rolls into a valley, its height decreases and its speed increases: potential energy turns into kinetic energy. When the car climbs the opposite slope, kinetic energy turns back into potential energy. If there is no friction or air resistance, the car rises exactly to its starting height and stops there for a moment. The mechanical energy is then equal at every moment, even though its form changes. This is called conservation of mechanical energy.
Not all the forces acting on the car change its energy. The weight does work when the car goes down or up, and it is exactly this work that turns potential energy into kinetic energy and back. The normal force of the track is perpendicular to the surface of the track and therefore also to the motion, so it does no work. That is why the mechanical energy is conserved when only the weight does work, however twisting the track is. The speed of the car at any point then depends only on how much lower it has descended from the starting point, not on the shape of the track or the mass of the car.
Work transfers energy
In the chapter Work and energy you learned that a force does work when an object moves in the direction of the force. Now we also need the sign of the work. A force does positive work when it has a component in the direction of the motion: it increases the energy of the object. A force that opposes the motion, such as friction, does negative work: it decreases the energy of the object. A force perpendicular to the motion, such as the normal force, does no work at all. The unit of work is the same as that of energy, the joule.
Friction turns mechanical energy into internal energy
In a real roller coaster the rolling resistance of the wheels and the air resistance oppose the motion the whole way. They do negative work, and the mechanical energy of the car decreases by exactly this work. Energy does not disappear but turns into internal energy of the wheels, the rails and the air, which warm up slightly. The total energy is always conserved. Mechanical energy is conserved only when there is no friction or air resistance, or when they are so small that they can be ignored.
You remember the difference between the weight, that is gravity, and friction from the chapter Kinetic energy, potential energy and mechanical energy: the weight is a conservative force. What is new is what follows from this for calculating energy: the work done by the weight depends only on the difference in height, so when an object returns to the same height, it gets back the same energy, whatever route it took. That is why a potential energy can be associated with the weight. The negative work done by friction and air resistance grows with every metre travelled, so it depends on the route. For such a non-conservative force a potential energy cannot be defined.
Answer to the opening question
A chain lift lifts the train to the top of the first hill. After that the train has only the mechanical energy it got during the lift, and nothing adds to it. The resistive forces decrease it with every metre. If the next hill were as high as the first, the train would not reach its top. Every later hill must be lower than the first and lower than the height that the remaining energy would reach, so that the train still has some kinetic energy at the top.
the work–energy principle
Work and kinetic energy
When an object moves in a straight line a distance $s$ and a constant force $\overline{F}$ acts on it, the work done by the force is
$$W = F_\parallel\, s$$
where $F_\parallel$ is the component of the force in the direction of the motion ($\text{N}$), negative when the component points against the motion, and $s$ is the distance travelled ($\text{m}$). The unit of work is $1\,\text{N}\cdot\text{m} = 1\,\text{J}$.
When a constant net force $\Sigma F$ in the direction of the motion acts on an object, it accelerates uniformly, and its acceleration is $a = \tfrac{\Sigma F}{m}$. In the chapter Acceleration and uniformly accelerated motion you learned that in uniformly accelerated motion the acceleration is $a = \tfrac{v_2 - v_1}{t}$ and the distance travelled is the average speed times the time, $s = \tfrac{v_1 + v_2}{2}\,t$. When these are multiplied together, the time cancels:
$$\begin{aligned} a s &= \frac{v_2 - v_1}{t} \cdot \frac{v_1 + v_2}{2}\,t \\ \phantom{as} &= \tfrac12 \left(v_2^2 - v_1^2\right) \end{aligned}$$
When the equation is multiplied on both sides by the mass $m$ and $ma = \Sigma F$ is substituted, we get $\Sigma F\, s = \tfrac{1}{2} m v_2^2 - \tfrac{1}{2} m v_1^2$. So the work done by the net force equals the change in kinetic energy:
$$W_\text{net} = \Delta E_\text{k}$$
where $W_\text{net}$ is the work done together by all the forces acting on the object ($\text{J}$) and $\Delta E_\text{k} = E_{\text{k},2} - E_{\text{k},1}$ is the change in kinetic energy ($\text{J}$). For example, when the mass of a cart is $2.0\,\text{kg}$ and a net force of $12\,\text{N}$ accelerates it from rest over a distance of $3.0\,\text{m}$, the force does the work $36\,\text{J}$, and the speed of the cart at the end is $6.0\,\text{m/s}$.
The work–energy principle
The total work is divided into two parts: the work done by the weight $W_G$ and the work done by all the other forces $W_\text{other}$. Other forces include, for example, friction, air resistance, a pulling rope and a pushing hand. When an object descends a height $h$, the weight does the work $mgh$ and the potential energy decreases by the same amount, so $W_G = -\Delta E_\text{p}$. Substituting this into the equation $W_G + W_\text{other} = \Delta E_\text{k}$ gives the work–energy principle of mechanics:
$$\begin{aligned} &E_{\text{k},1} + E_{\text{p},1} + W_\text{other} \\ \quad &= E_{\text{k},2} + E_{\text{p},2} \end{aligned}$$
where the index $1$ means the initial situation and $2$ the final situation, $E_\text{k} = \tfrac{1}{2} mv^2$ is the kinetic energy, $E_\text{p} = mgh$ is the potential energy at a height $h$ from the chosen zero level, and $W_\text{other}$ is the work done by forces other than the weight ($\text{J}$). In words: the change in mechanical energy equals the work done by forces other than the weight, $\Delta E_\text{mech} = W_\text{other}$. The zero level can be chosen freely, for example at the lowest point of the situation, because only the change in height affects the result.
When an object moves against a stationary surface or air, friction and air resistance oppose the motion and do negative work, so they decrease the mechanical energy. (A moving surface or the wind can also do positive work, for example a conveyor belt carrying a box.) A pulling rope or a motor can do positive work and increase it, like the chain lift that lifts the train. If the other forces do no work, $W_\text{other} = 0$, and the work–energy principle becomes the law of conservation of mechanical energy:
$$\begin{aligned} &E_{\text{k},1} + E_{\text{p},1} \\ \quad &= E_{\text{k},2} + E_{\text{p},2} \end{aligned}$$
When an object starts from rest at a height $h$ and descends to the zero level, the conservation law $mgh = \tfrac{1}{2} mv^2$ gives $v = \sqrt{2gh}$. The mass cancels, so the speed does not depend on the mass.
Common misconception
Many people think that a heavier object is faster at the bottom of a hill. When friction and air resistance can be ignored, the mass cancels from the conservation law: a full and an empty car are equally fast at the bottom of the valley. The heavier car has more kinetic energy at the bottom, but it also had more potential energy at the start.
Another mistake is to say that friction destroys energy. Friction decreases the mechanical energy, but an equal amount of energy turns into internal energy, and the total energy is conserved. A third mistake is to calculate the work of friction from the difference in height. The work of friction is the frictional force times the distance travelled along the track. When the resistive force is the same, on a long gentle track more mechanical energy turns into internal energy than on a short steep one, even if the difference in height is the same.
Summary
- The mechanical energy $E_\text{mech} = E_\text{k} + E_\text{p}$ is conserved when only the weight does work, that is, when friction and air resistance can be ignored and nothing pulls or pushes. The normal force does no work, because it is perpendicular to the motion.
- A force does work with its component in the direction of the motion, $W = F_\parallel s$: positive in the direction of the motion and negative against it.
- The work done by the net force equals the change in kinetic energy, $W_\text{net} = \Delta E_\text{k}$.
- The work–energy principle: the change in mechanical energy is the work done by forces other than the weight, $\Delta E_\text{mech} = W_\text{other}$. Friction and air resistance turn mechanical energy into internal energy, and the total energy is conserved.
- The weight is a conservative force, because its work depends only on the difference in height. Friction and air resistance are non-conservative forces, because their work grows with the distance travelled.
Optional Return to your predictions
Compare your predictions with the readings of the simulation. What surprised you most? Write in your own words what decides the speed of the car at the bottom of the valley and what decides how much mechanical energy turns into internal energy.
Would you change your prediction?
Apply
Worked example
Problem
A child slides down a hill on a sled. The combined mass of the child and the sled is $32\,\text{kg}$. The sled starts from rest at the top of the hill, which is $5.0\,\text{m}$ higher than the foot of the hill. At the foot of the hill the speed of the sled is measured to be $7.6\,\text{m/s}$. How much work do friction and air resistance together do on the sled during the descent?
Step 1/7
Given and find. $m = 32\,\text{kg}$, difference in height $h = 5.0\,\text{m}$, initial speed $v_1 = 0$, final speed $v_2 = 7.6\,\text{m/s}$ and $g = 9.81\,\text{m/s}^2$. We want the work $W_\text{other}$ done together by friction and air resistance.
Step 2/7
Principle. The weight, the normal force of the snow, friction and air resistance act on the sled. The normal force is perpendicular to the motion, so it does no work. The work of the weight is taken into account as the change in potential energy. Friction and air resistance are other forces, so we use the work–energy principle. The zero level is chosen at the foot of the hill.
Step 3/7
Formula solved. At the start $E_{\text{k},1} = 0$ and $E_{\text{p},1} = mgh$, at the end $E_{\text{k},2} = \tfrac{1}{2} mv_2^2$ and $E_{\text{p},2} = 0$:
$$\begin{aligned} &mgh + W_\text{other} = \tfrac{1}{2} m v_2^2 \\ &W_\text{other} = \tfrac{1}{2} m v_2^2 - mgh \end{aligned}$$
Step 4/7
Unit conversions. Not needed: all quantities are in SI units, and $1\,\text{kg}\cdot\tfrac{\text{m}^2}{\text{s}^2} = 1\,\text{J}$.
Step 5/7
Substitution. First we calculate the kinetic energy at the end and the potential energy at the start, and then their difference:
$$\begin{aligned} &\tfrac{1}{2} m v_2^2 \\ &= \tfrac12 \cdot 32\,\text{kg} \cdot \left(7.6\,\tfrac{\text{m}}{\text{s}}\right)^2 \\ &= 924.16\,\text{J} \\[6pt] &mgh \\ &= 32\,\text{kg} \cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \cdot 5.0\,\text{m} \\ &= 1569.6\,\text{J} \\[6pt] &W_\text{other} \\ &= 924.16\,\text{J} - 1569.6\,\text{J} \\ &= -645.44\,\text{J} \end{aligned}$$
Step 6/7
Result. Friction and air resistance together do the work $W_\text{other} \approx -6.5 \cdot 10^2\,\text{J}$. The result is given with two significant figures like the height $5.0\,\text{m}$ and the speed $7.6\,\text{m/s}$. The work is negative, because friction and air resistance oppose the motion: about $650\,\text{J}$ of mechanical energy turned into internal energy of the snow, the runners and the air.
Step 7/7
Reasonableness. Without friction and air resistance the speed of the sled at the foot of the hill would be $\sqrt{2gh} \approx 9.9\,\text{m/s}$, so the measured $7.6\,\text{m/s}$ is smaller, as it should be. About $41\,\%$ of the initial potential energy turned into internal energy. If the slope is $24\,\text{m}$ long, the average resistive force is $645\,\text{J} : 24\,\text{m} \approx 27\,\text{N}$, less than a tenth of the weight $314\,\text{N}$, which is believable for a sliding sled.
Optional Find the error
Apply to a new situation
A child sits on a swing at a playground. The swing is pulled back so that the seat is $1.20\,\text{m}$ higher than at the lowest point of the swing, and it is released from rest. Air resistance and the friction at the attachment of the chains are ignored.
What is the speed of the seat at the lowest point of the swing? Answer in $\text{m/s}$.
Hint 1: a guiding question
Which forces act on the seat and the child, and which of them do work? The tension force of the chains is perpendicular to the motion at every moment.
Hint 2: what is the connection
Only the weight does work, so the mechanical energy is conserved: $mgh = \tfrac{1}{2} mv^2$. Choose the zero level at the lowest point.
Hint 3: an intermediate step
The mass cancels:
Show the solution
Given
difference in height $h = 1.20\,\text{m}$, initial speed $v_1 = 0$, $g = 9.81\,\text{m/s}^2$; air resistance and friction are ignored
Find
the speed $v$ at the lowest point in $\text{m/s}$
Principle
The weight and the tension force of the chains act on the seat and the child. The seat moves along a circular arc, and the force of the chains is perpendicular to the motion at every moment, so it does no work. Only the weight does work, so the mechanical energy is conserved. The zero level is chosen at the lowest point.
Formula
$mgh = \tfrac{1}{2} mv^2$, so $v = \sqrt{2gh}$
Conversions
The height is in metres, so no conversions are needed. The mass is not needed, because it cancels.
Substitution
Answer
$4.85\,\text{m/s}$
Three significant figures like the height $1.20\,\text{m}$.
Reasonableness
The speed is about $17\,\text{km/h}$, which suits brisk swinging. The same speed would be reached by dropping an object straight down from a height of $1.20\,\text{m}$: the curved path of the swing does not change the speed, because the work done by the weight depends only on the difference in height.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A box slides on a horizontal floor. Which of the forces acting on the box do work, and which do not? Give reasons.
Answer
Friction does negative work, because it points against the motion. The weight and the normal force of the floor are perpendicular to the motion, so they do no work.
- A sled slides on level snow and stops. Where did its kinetic energy go, and was energy conserved?
Answer
Friction and air resistance did negative work, and the kinetic energy turned into internal energy of the snow, the runners and the air. The mechanical energy decreased, but the total energy was conserved.
1. Concept
2. Calculating
A roller coaster car starts from rest at the top of a hill at a height of $18.0\,\text{m}$ and goes over the top of the next hill, which is at a height of $7.0\,\text{m}$. Friction and air resistance are ignored.
What is the speed of the car at the top of the second hill? Answer in $\text{m/s}$.
Hint 1: a guiding question
Does any force other than the weight do work? If not, the mechanical energy is conserved.
Hint 2: what is the connection
Choose the zero level at the ground: $mgh_1 = \tfrac{1}{2} mv^2 + mgh_2$.
Hint 3: an intermediate step
The mass cancels: $v = \sqrt{2g(h_1 - h_2)}$, and the difference in height is $11.0\,\text{m}$.
Show the solution
Given
starting height $h_1 = 18.0\,\text{m}$, top of the second hill $h_2 = 7.0\,\text{m}$, initial speed $v_1 = 0$, $g = 9.81\,\text{m/s}^2$; friction and air resistance are ignored
Find
the speed $v_2$ at the top of the second hill in $\text{m/s}$
Principle
The weight and the normal force of the track act on the car. The normal force is perpendicular to the motion and does no work, so only the weight does work and the mechanical energy is conserved. The zero level is chosen at the ground.
Formula
$mgh_1 = \tfrac{1}{2} mv_2^2 + mgh_2$, so $v_2 = \sqrt{2g(h_1 - h_2)}$
Conversions
The heights are in metres, so no conversions are needed. The mass is not needed, because it cancels.
Substitution
Answer
$14.7\,\text{m/s}$
Three significant figures like the difference in height $11.0\,\text{m}$.
Reasonableness
The speed is about $53\,\text{km/h}$. It is smaller than the speed $\sqrt{2g \cdot 18.0\,\text{m}} \approx 18.8\,\text{m/s}$ at ground level, as it should be, because at the top the car still has potential energy.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Explain
Use energy to explain why the first hill of a roller coaster is the highest of the track and the later hills are lower.
Where does the mechanical energy of the train go during the ride?
Model answer
The chain lift does work on the train and lifts it to the top of the first hill. After that only the weight, the normal force of the track and the resistive forces act on the train, and none of them adds to the mechanical energy of the train. If there were no resistive forces, the mechanical energy would be conserved and the train could rise at most to the starting height, and even then it would stop at the top. In reality rolling resistance and air resistance do negative work all the time, and it grows with every metre travelled. So the mechanical energy of the train decreases during the whole ride. Each hill must be lower than the height that the remaining mechanical energy would reach, so that the train still has kinetic energy at the top. That is why the hills usually get lower towards the end of the track. Energy does not disappear: it turns into internal energy of the wheels, the rails and the air, and finally at the station the brakes turn the rest of the kinetic energy into internal energy.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Review the chapter's concepts
Cards you mark Review later collect in your review deck.
- Mechanical work
- A force transfers energy to or from an object during a displacement.$$W = Fs\cos\alpha$$
- Conservative force
- The work done by the force depends only on the start and end positions, not on the path.
- Conservation of energy
- In an isolated system the total energy stays constant.
- Conservation of mechanical energy
- Mechanical energy is conserved when only conservative forces do work.
- Work–energy principle of mechanics
- The change in mechanical energy equals the work done by the non-conservative forces.$$\Delta E_{\text{mech}} = W_{\text{other}}$$
- Work–kinetic energy principle
- The work done by the net force on an object equals the change in its kinetic energy.$$W_{\text{total}} = \Delta E_{\text{k}}$$
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 11-1Core
A stone is dropped from rest from a bridge $12.0\,\text{m}$ above the water. Air resistance is ignored. At what speed does the stone hit the water?
Answer
$15.3\,\text{m/s}$
Show the solution
Given
drop height $h = 12.0\,\text{m}$, initial speed $v_1 = 0$, $g = 9.81\,\text{m/s}^2$; air resistance is ignored
Find
the speed $v$ at the water surface in $\text{m/s}$
Principle
Only the weight acts on the stone, so only the weight does work and the mechanical energy is conserved. The zero level is chosen at the water surface: at the start the stone has only potential energy, at the end only kinetic energy.
Formula
$mgh = \tfrac{1}{2} mv^2$, so $v = \sqrt{2gh}$
Conversions
The height is in metres, so no conversions are needed. The mass is not needed, because it cancels.
Substitution
$v$$= \sqrt{2 \cdot 9.81\,\text{m/s}^2 \cdot 12.0\,\text{m}}$$= \sqrt{235.44\,\text{m}^2/\text{s}^2}$$= 15.34\,\text{m/s}$Answer
$15.3\,\text{m/s}$
Three significant figures like the height $12.0\,\text{m}$.
Reasonableness
Estimate with round numbers:
$\sqrt{2 \cdot 10 \cdot 12}\,\text{m/s}$$= \sqrt{240}\,\text{m/s}$$\approx 15.5\,\text{m/s}$so the result is of the right order of magnitude. The speed is about $55\,\text{km/h}$, which suits a drop from the height of a four-storey building.
- Exercise 11-2Core
A skateboarder moves at the bottom of a ramp at a speed of $6.0\,\text{m/s}$ and rises up the other side of the ramp without pushing. Friction and air resistance are ignored.
How high above the bottom of the ramp does the skateboarder rise?
Answer
$1.8\,\text{m}$
Show the solution
Given
speed at the bottom $v = 6.0\,\text{m/s}$, speed at the highest point $0$, $g = 9.81\,\text{m/s}^2$; friction and air resistance are ignored
Find
the rise $h$ above the bottom of the ramp in $\text{m}$
Principle
The weight and the normal force of the ramp act on the skateboarder. The normal force is perpendicular to the motion and does no work, so the mechanical energy is conserved. The zero level is at the bottom of the ramp: the kinetic energy turns entirely into potential energy when the skateboarder stops at the highest point.
Formula
$\tfrac{1}{2} mv^2 = mgh$, so $h = \dfrac{v^2}{2 \cdot g}$
Conversions
The speed is in $\text{m/s}$, so no conversions are needed. The mass cancels.
Substitution
$h$$= \dfrac{(6.0\,\text{m/s})^2}{2 \cdot 9.81\,\text{m/s}^2}$$= \dfrac{36\,\text{m}^2/\text{s}^2}{19.62\,\text{m/s}^2}$$= 1.835\,\text{m}$Answer
$1.8\,\text{m}$
Two significant figures like the speed $6.0\,\text{m/s}$.
Reasonableness
When falling $1.8\,\text{m}$, the speed is $\sqrt{2 \cdot 9.81 \cdot 1.835}\,\text{m/s} = 6.0\,\text{m/s}$, so the calculation works backwards too. In reality friction and air resistance reduce the rise slightly.
- Exercise 11-3
A box is pushed a distance of $4.0\,\text{m}$ on a horizontal floor. The frictional force of the floor on the box is $35\,\text{N}$.
How much work does the frictional force do on the box? Give the work with its sign.
Answer
$-140\,\text{J}$
Show the solution
Given
distance $s = 4.0\,\text{m}$, frictional force $F_\mu = 35\,\text{N}$ against the motion
Find
the work $W$ done by the frictional force, with its sign, in $\text{J}$
Principle
The frictional force points against the motion, so its component in the direction of the motion is $-F_\mu$. A force that opposes the motion does negative work.
Formula
$W = F_\parallel s$, and $F_\parallel = -F_\mu$, so $W = -F_\mu s$
Conversions
The force is in newtons and the distance in metres, so no conversions are needed; $1\,\text{N}\cdot\text{m} = 1\,\text{J}$.
Substitution
$W$$= -35\,\text{N} \cdot 4.0\,\text{m}$$= -140\,\text{J}$Answer
$-140\,\text{J}$
That is $-1.4 \cdot 10^2\,\text{J}$, two significant figures like the distance $4.0\,\text{m}$ and the force $35\,\text{N}$.
Reasonableness
The work is negative, because friction decreases the energy of the box: $140\,\text{J}$ turns into internal energy of the box and the floor. Order of magnitude: about $40\,\text{N}$ times $4\,\text{m}$ is about $160\,\text{J}$.
- Exercise 11-4
A sprinter of mass $65\,\text{kg}$ accelerates from rest at the start to a speed of $8.0\,\text{m/s}$.
How much work does the net force on the sprinter do during the acceleration? Answer in kilojoules.
Answer
$2.1\,\text{kJ}$
Show the solution
Given
$m = 65\,\text{kg}$, initial speed $v_1 = 0$, final speed $v_2 = 8.0\,\text{m/s}$
Find
the work $W_\text{net}$ done by the net force in $\text{kJ}$
Principle
The work done by the net force equals the change in kinetic energy. The net force or the distance need not be known, because the initial and final speeds tell the change in kinetic energy.
Formula
$W_\text{net}$$= \Delta E_\text{k}$$= \tfrac{1}{2} mv_2^2 - \tfrac{1}{2} mv_1^2$and $v_1 = 0$, so $W_\text{net} = \tfrac{1}{2} mv_2^2$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$W_\text{net}$$= \tfrac12 \cdot 65\,\text{kg} \cdot (8.0\,\text{m/s})^2$$= 2080\,\text{J}$Answer
$2.1\,\text{kJ}$
Two significant figures like the mass $65\,\text{kg}$ and the speed $8.0\,\text{m/s}$.
Reasonableness
The work is positive, because the kinetic energy of the sprinter increases. Estimate: $\tfrac12 \cdot 60 \cdot 64\,\text{J} \approx 1900\,\text{J}$, so the result is of the right order of magnitude.
- Exercise 11-5
The mechanical energy of a skier is $12.4\,\text{kJ}$ at the top of a hill and $9.8\,\text{kJ}$ at the bottom.
How much work do friction and air resistance together do on the skier during the descent? Give the work with its sign.
Answer
$-2.6\,\text{kJ}$
Show the solution
Given
mechanical energy at the top of the hill $E_{\text{mech},1} = 12.4\,\text{kJ}$ and at the bottom $E_{\text{mech},2} = 9.8\,\text{kJ}$
Find
the work $W_\text{other}$ done together by friction and air resistance, with its sign, in $\text{kJ}$
Principle
By the work–energy principle the change in mechanical energy equals the work done by forces other than the weight. Here the other forces are friction and air resistance; the normal force of the snow does no work, because it is perpendicular to the motion.
Formula
$E_{\text{mech},1} + W_\text{other} = E_{\text{mech},2}$, so $W_\text{other} = E_{\text{mech},2} - E_{\text{mech},1}$
Conversions
Both energies are in kilojoules, so the answer comes directly in kilojoules.
Substitution
$W_\text{other}$$= 9.8\,\text{kJ} - 12.4\,\text{kJ}$$= -2.6\,\text{kJ}$Answer
$-2.6\,\text{kJ}$
Two significant figures, because the energies are known to a tenth of a kilojoule and their difference is $2.6\,\text{kJ}$.
Reasonableness
The work is negative, because friction and air resistance oppose the motion and the mechanical energy decreases. The skier lost about a fifth of the mechanical energy, which is believable in downhill skiing.
- Exercise 11-6
Two children slide down slides from rest. One slide is steep and short, the other gentle and long, but the starting height of both is the same. The children have equal masses, and friction and air resistance are ignored.
What can be concluded about the children’s speeds at the bottom of the slide?
Answer
The speeds are equal, because the change in potential energy is the same.
Show the solution
Find
how the speeds of the children at the bottom of the slide compare with each other
Principle
The weight and the normal force of the slide act on each child. The normal force is perpendicular to the motion and does no work, and friction and air resistance are ignored, so the mechanical energy is conserved. The work done by the weight, that is the change in potential energy, depends only on the height difference, not on the shape or the length of the track.
Model answer
Both children start from rest at the same height and have the same mass, so the initial potential energy $mgh$ is the same. At the bottom it has turned entirely into kinetic energy, $mgh = \tfrac{1}{2} mv^2$, so $v = \sqrt{2gh}$ is the same for both. On the steep slide the acceleration is greater but the distance is shorter, and on the gentle slide the acceleration is smaller but the distance is longer; these balance each other, and the speeds at the bottom are equal. Only the time taken is different.
- Exercise 11-7
A skater of mass $58.0\,\text{kg}$ glides on ice at a speed of $7.50\,\text{m/s}$. Friction and air resistance slow the motion, and the skater’s speed decreases to $4.50\,\text{m/s}$.
How much work does the net force on the skater do during the slowing down? Give the work with its sign in kilojoules.
Answer
$-1.04\,\text{kJ}$
Show the solution
Given
$m = 58.0\,\text{kg}$, initial speed $v_1 = 7.50\,\text{m/s}$, final speed $v_2 = 4.50\,\text{m/s}$
Find
the work $W_\text{net}$ done by the net force, with its sign, in $\text{kJ}$
Principle
The work done by the net force equals the change in kinetic energy. The net force and the distance do not need to be known, because the initial and final speeds give the change in kinetic energy.
Formula
$W_\text{net}$$= \Delta E_\text{k}$$= \tfrac{1}{2} mv_2^2 - \tfrac{1}{2} mv_1^2$Conversions
The quantities are in SI units. Finally $1\,\text{kJ} = 1000\,\text{J}$.
Substitution
First
$v_2^2 - v_1^2$$= (20.25 - 56.25)\,\text{m}^2/\text{s}^2$$= -36.0\,\text{m}^2/\text{s}^2$so
$W_\text{net}$$= \tfrac12 \cdot 58.0\,\text{kg} \cdot (-36.0\,\text{m}^2/\text{s}^2)$$= -1044\,\text{J}$$= -1.044\,\text{kJ}$Answer
$-1.04\,\text{kJ}$
Three significant figures like the mass $58.0\,\text{kg}$ and the speeds.
Reasonableness
The work is negative, because friction and air resistance oppose the motion and the kinetic energy decreases. At the start the kinetic energy is $\tfrac12 \cdot 58 \cdot 7.50^2\,\text{J} \approx 1630\,\text{J}$ and at the end $\tfrac12 \cdot 58 \cdot 4.50^2\,\text{J} \approx 590\,\text{J}$; the difference is about $-1040\,\text{J}$.
- Exercise 11-8
A student solves the problem: “A box slides down an inclined plank a distance of $5.0\,\text{m}$. The height difference between the ends of the plank is $1.5\,\text{m}$, and the frictional force opposing the motion is $8.0\,\text{N}$.
How much work does friction do on the box?”
The student’s solution: $W = -F_\mu h = -8.0\,\text{N} \cdot 1.5\,\text{m} = -12\,\text{J}$.
What is wrong in the student’s solution?
Answer
The work of friction is calculated with the distance travelled and not with the height difference.
Show the solution
Find
what is wrong in the student's solution
Principle
The work done by a force is $W = F_\parallel s$, where $s$ is the distance travelled in the direction of the force. The frictional force points against the motion all the way, so the work of friction is $-F_\mu s$, where $s$ is the distance along the plank. The height difference belongs to the work of the weight, that is the change in potential energy, not to the work of friction.
Model answer
The student has multiplied the frictional force by the height difference between the ends of the plank, although the work of friction needs the distance the box travels along the plank. The frictional force acts along the whole length of the plank, so the correct work is $W = -8.0\,\text{N} \cdot 5.0\,\text{m} = -40\,\text{J}$. The sign $-$ was correct, because friction opposes the motion; only the distance was wrong. The student's result $-12\,\text{J}$ is too small.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 11-9Core
A ball is thrown straight down from a balcony at a height of $15.0\,\text{m}$ at a speed of $8.00\,\text{m/s}$. Air resistance is ignored.
At what speed does the ball hit the ground?
Answer
$18.9\,\text{m/s}$
Show the solution
Given
height $h = 15.0\,\text{m}$, initial speed $v_1 = 8.00\,\text{m/s}$ downward, $g = 9.81\,\text{m/s}^2$; air resistance is ignored
Find
the speed $v_2$ at the ground in $\text{m/s}$
Principle
After the throw only the weight acts on the ball, so the mechanical energy is conserved. The zero level is chosen at the ground: at the start the ball has both kinetic and potential energy, at the end only kinetic energy. Kinetic energy does not depend on the direction of the velocity.
Formula
$\tfrac{1}{2} mv_1^2 + mgh = \tfrac{1}{2} mv_2^2$, so $v_2 = \sqrt{v_1^2 + 2gh}$
Conversions
The quantities are in SI units, so no conversions are needed. The mass cancels.
Substitution
First
$v_1^2$$= (8.00\,\text{m/s})^2$$= 64.0\,\text{m}^2/\text{s}^2$$2gh$$= 2 \cdot 9.81\,\text{m/s}^2 \cdot 15.0\,\text{m}$$= 294.3\,\text{m}^2/\text{s}^2$so
$v_2$$= \sqrt{(64.0 + 294.3)\,\text{m}^2/\text{s}^2}$$= 18.93\,\text{m/s}$Answer
$18.9\,\text{m/s}$
Three significant figures like the height $15.0\,\text{m}$ and the speed $8.00\,\text{m/s}$.
Reasonableness
A plain drop would give $\sqrt{294.3}\,\text{m/s} \approx 17.2\,\text{m/s}$, and the throw adds speed, so the result must be greater than this. The speeds do not simply add, however: $18.9\,\text{m/s}$ is much less than $8.0 + 17.2\,\text{m/s}$, because energies add, not speeds.
- Exercise 11-10Core
A car of mass $1200\,\text{kg}$ drives on a horizontal road at a speed of $72\,\text{km/h}$. The driver brakes, and the frictional force of the road on the tyres is $6.0\,\text{kN}$ during the braking. Air resistance is ignored.
How long is the braking distance of the car?
Answer
$40\,\text{m}$
Show the solution
Given
$m = 1200\,\text{kg}$, initial speed $v_1 = 72\,\text{km/h}$, final speed $v_2 = 0$, frictional force $F_\mu = 6.0\,\text{kN}$ against the motion; the road is horizontal and air resistance is ignored
Find
the braking distance $s$ in $\text{m}$
Principle
The road is horizontal, so the potential energy does not change. The weight and the normal force of the road are perpendicular to the motion and do no work. The only force that does work is friction, which is also the net force, so its work is the change in kinetic energy.
Formula
$-F_\mu s = 0 - \tfrac{1}{2} mv_1^2$, so $s = \dfrac{mv_1^2}{2F_\mu}$
Conversions
$v_1$$= 72\,\text{km/h}$$= 72 : 3.6\,\text{m/s}$$= 20\,\text{m/s}$$F_\mu$$= 6.0\,\text{kN}$$= 6.0 \cdot 1000\,\text{N}$$= 6000\,\text{N}$Substitution
$s$$= \dfrac{1200\,\text{kg} \cdot (20\,\text{m/s})^2}{2 \cdot 6000\,\text{N}}$$= \dfrac{480\,000\,\text{J}}{12\,000\,\text{N}}$$= 40\,\text{m}$Answer
$40\,\text{m}$
Two significant figures like the speed $72\,\text{km/h}$ and the force $6.0\,\text{kN}$.
Reasonableness
The deceleration is $6000\,\text{N} : 1200\,\text{kg} = 5.0\,\text{m/s}^2$, and the equations of motion give the same distance: $\tfrac{(20\,\text{m/s})^2}{2 \cdot 5.0\,\text{m/s}^2} = 40\,\text{m}$. Kinetic energy is proportional to the square of the speed, so at double the speed the braking distance would be four times as long.
- Exercise 11-11
A crane lifts a $250\,\text{kg}$ concrete load $18.0\,\text{m}$ upward at a constant speed. How much work does the rope of the crane do on the load? Answer in kilojoules.
Answer
$44.1\,\text{kJ}$
Show the solution
Given
$m = 250\,\text{kg}$, lifting height $h = 18.0\,\text{m}$, speed constant, $g = 9.81\,\text{m/s}^2$
Find
the work $W_\text{other}$ done by the rope in $\text{kJ}$
Principle
The weight and the tension force of the rope act on the load. The rope is a force other than the weight, so we use the work–energy principle. The speed is constant, so the kinetic energy does not change, and the work of the rope increases only the potential energy. The zero level is chosen at the starting height of the load.
Formula
$E_{\text{k},1} + E_{\text{p},1} + W_\text{other}$$= E_{\text{k},2} + E_{\text{p},2}$and $E_{\text{k},1} = E_{\text{k},2}$ and $E_{\text{p},1} = 0$, so $W_\text{other} = mgh$
Conversions
The quantities are in SI units. Finally $1\,\text{kJ} = 1000\,\text{J}$.
Substitution
$W_\text{other}$$= 250\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 18.0\,\text{m}$$= 44\,145\,\text{J}$$= 44.1\,\text{kJ}$Answer
$44.1\,\text{kJ}$
Three significant figures like the lifting height $18.0\,\text{m}$ and the mass $250\,\text{kg}$.
Reasonableness
At a constant speed the force of the rope equals the weight, about $2.5\,\text{kN}$, and $2.5\,\text{kN} \cdot 18\,\text{m} = 45\,\text{kJ}$. The work is positive, because the rope pulls in the direction of the motion and increases the mechanical energy of the load.
- Exercise 11-12
An ice hockey player hits a puck of mass $0.16\,\text{kg}$ that is at rest. The stick pushes the puck in the direction of motion with an average force of $120\,\text{N}$ over a distance of $0.30\,\text{m}$. The friction of the ice is ignored.
What is the speed of the puck when it leaves the stick?
Answer
$21\,\text{m/s}$
Show the solution
Given
$m = 0.16\,\text{kg}$, initial speed $v_1 = 0$, force of the stick $F = 120\,\text{N}$ in the direction of the motion, distance $s = 0.30\,\text{m}$; the friction of the ice is ignored
Find
the speed $v_2$ of the puck at the moment it leaves the stick in $\text{m/s}$
Principle
The weight, the normal force of the ice and the force of the stick act on the puck. The weight and the normal force cancel each other and do no work in horizontal motion, so the force of the stick is the net force. The work done by the net force equals the change in kinetic energy.
Formula
$Fs = \tfrac{1}{2} mv_2^2 - 0$, so $v_2 = \sqrt{\dfrac{2Fs}{m}}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$v_2$$= \sqrt{\dfrac{2 \cdot 120\,\text{N} \cdot 0.30\,\text{m}}{0.16\,\text{kg}}}$$= \sqrt{\dfrac{72\,\text{J}}{0.16\,\text{kg}}}$$= 21.2\,\text{m/s}$Answer
$21\,\text{m/s}$
Two significant figures like the mass $0.16\,\text{kg}$ and the distance $0.30\,\text{m}$.
Reasonableness
The speed is about $76\,\text{km/h}$, which is a normal hard shot. The work $36\,\text{J}$ done by the stick equals the kinetic energy of the puck:
$\tfrac12 \cdot 0.16\,\text{kg} \cdot (21.2\,\text{m/s})^2$$\approx 36\,\text{J}$ - Exercise 11-13
Students study the energy of a pendulum with a light gate. The bob of the pendulum is released from rest at different heights $h$ measured from the lowest point of the pendulum, and a light gate at the lowest point measures the speed $v$ of the bob. Friction and air resistance are ignored. The table is example data.
a) Calculate $v^2$ for each measurement, plot the points in an $(h, v^2)$ coordinate system and fit a straight line through the origin to them. Determine the slope of the line.
b) Derive from the conservation law the relation between the slope and the acceleration of free fall $g$, and calculate $g$.
c) Justify why the line must pass through the origin.
The answer field takes the result of part b).
Speed of the bob at the lowest point of the pendulum (example data) $h\,(\text{m})$ $v\,(\text{m/s})$ 0.05 1.00 0.10 1.38 0.15 1.73 0.20 1.96 0.25 2.23 0.30 2.41 Answer
$g \approx 9.8\,\text{m/s}^2$
Show the solution
Given
the height $h$ of the bob from the lowest point and the speed $v$ at the lowest point in six measurements of the table; initial speed $0$; friction and air resistance are ignored
Find
the acceleration of free fall $g$ in $\text{m/s}^2$
Principle
The weight and the tension of the string act on the bob. The tension is perpendicular to the motion and does no work, so the mechanical energy is conserved. The zero level is the lowest point: at the top the bob has only potential energy, at the bottom only kinetic energy. The mass cancels, so the mass of the bob is not needed.
Reading the graph
The squares are $v^2 = 1.00$; $1.90$; $2.99$; $3.84$; $4.97$; $5.81\,\text{m}^2/\text{s}^2$. The points $(h, v^2)$ lie close to a straight line. The line passes through the origin, so its slope is obtained from one point read from the line: the line passes between the points through the point $(0.30\,\text{m}$; $5.86\,\text{m}^2/\text{s}^2)$, so the slope is
$k$$= 5.86\,\text{m}^2/\text{s}^2 : 0.30\,\text{m}$$= 19.5\,\text{m/s}^2$Formula
$mgh = \tfrac{1}{2} mv^2$, so $v^2 = 2g \cdot h$. The slope of the line is $k = 2g$, so $g = \dfrac{k}{2}$.
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$g$$= \dfrac{19.5\,\text{m/s}^2}{2}$$= 9.77\,\text{m/s}^2$Answer
$g \approx 9.8\,\text{m/s}^2$
Two significant figures like the heights $h$ (for example $0.05\,\text{m}$). The more precise value read from the graph is $9.77\,\text{m/s}^2$.
Reasonableness
The result differs from the known value $9.81\,\text{m/s}^2$ by less than one percent, which is within the measurement error. A single measurement gives the same order of magnitude: $h = 0.30\,\text{m}$ and $v = 2.41\,\text{m/s}$ give $g = v^2 : (2h) = 9.68\,\text{m/s}^2$. c) The line must pass through the origin, because when $h = 0$ the bob is at rest at the lowest point and has no kinetic energy, so $v = 0$ and $v^2 = 0$. The equation $v^2 = 2gh$ is a proportionality with no constant term.
- Exercise 11-14
A mini roller coaster car with its passengers has a mass of $420\,\text{kg}$. The height $h$ and speed $v$ of the car were measured at four points of the track, and $s$ is the track distance from point A. The table is example data. The zero level is the ground.
a) Calculate the mechanical energy of the car at points A, B, C and D in kilojoules.
b) How much work do the resistive forces do on the car in total between points A and D?
c) Between which points is the average resistive force greatest? Justify by calculating.
The answer field takes the result of part b) in kilojoules.
Measurements of the car at different points of the track (example data) Point $s\,(\text{m})$ $h\,(\text{m})$ $v\,(\text{m/s})$ A 0 18.0 1.5 B 55 6.0 14.6 C 120 12.0 7.9 D 170 3.0 14.2 Answer
$W_\text{other} \approx -19.9\,\text{kJ}$
Show the solution
Given
$m = 420\,\text{kg}$, $g = 9.81\,\text{m/s}^2$; at points A, B, C, D the track distance $s = 0$; $55$; $120$; $170\,\text{m}$, the height $h = 18.0$; $6.0$; $12.0$; $3.0\,\text{m}$ and the speed $v = 1.5$; $14.6$; $7.9$; $14.2\,\text{m/s}$
Find
a) the mechanical energy $E_\text{mech}$ at points A–D; b) the work $W_\text{other}$ of the resistive forces between points A and D; c) the section between points where the average resistive force is greatest
Principle
The mechanical energy is $E_\text{mech} = \tfrac{1}{2} mv^2 + mgh$ with the ground as the zero level. The normal force does no work, so the change in the mechanical energy of the car is the work of the resistive forces, $W_\text{other} = E_{\text{mech},2} - E_{\text{mech},1}$. The resistive force points against the motion all the way, so its average over a section is $F = -W_\text{other} : \Delta s$.
Formula
$E_\text{mech} = \tfrac{1}{2} mv^2 + mgh$ and $W_\text{other} = E_{\text{mech,D}} - E_{\text{mech,A}}$
Conversions
The quantities are in SI units. Finally $1\,\text{kJ} = 1000\,\text{J}$.
Substitution
a) At A it is
$\tfrac12 \cdot 420 \cdot 1.5^2\,\text{J} + 420 \cdot 9.81 \cdot 18.0\,\text{J}$$= 0.47\,\text{kJ} + 74.16\,\text{kJ}$$= 74.6\,\text{kJ}$In the same way the kinetic energies at B, C and D are $44.8$; $13.1$; $42.3\,\text{kJ}$ and the potential energies $24.7$; $49.4$; $12.4\,\text{kJ}$, so $E_\text{mech} = 69.5$; $62.5$; $54.7\,\text{kJ}$. b)
$W_\text{other}$$= 54.7\,\text{kJ} - 74.6\,\text{kJ}$$= -19.9\,\text{kJ}$c) Between A and B the energy decreases by $5.15\,\text{kJ}$ over $55\,\text{m}$, so
$F$$= 5150\,\text{J} : 55\,\text{m}$$= 94\,\text{N}$Between B and C it decreases by $6.94\,\text{kJ}$ over $65\,\text{m}$, so $F = 107\,\text{N}$. Between C and D it decreases by $7.84\,\text{kJ}$ over $50\,\text{m}$, so $F = 157\,\text{N}$.
Answer
b) $W_\text{other} \approx -19.9\,\text{kJ}$, three significant figures like the heights and speeds. c) The average resistive force is greatest between points C and D, about $1.6 \cdot 10^2\,\text{N}$.
Reasonableness
The mechanical energy decreases in every section, as it should, and altogether about $27\,\%$ of the initial energy turned into internal energy. The average resistive force over the whole track is $19\,900\,\text{J} : 170\,\text{m} \approx 117\,\text{N}$, which lies between the averages of the three sections, $94\,\text{N}$, $107\,\text{N}$ and $157\,\text{N}$, as it should.
- Exercise 11-15
Students determine the frictional force by an experiment. A wooden block of mass $0.50\,\text{kg}$ is allowed to slide on a horizontal table with different initial speeds $v_0$, and each time the stopping distance $s$ of the block is measured. Air resistance is ignored and the frictional force is assumed to be constant. The table is example data.
a) Calculate $v_0^2$, plot the points in a $(v_0^2, s)$ coordinate system and fit a straight line through the origin to them. Determine the slope of the line.
b) Derive from the work–energy principle the relation between the slope and the frictional force $F_\mu$, and calculate the frictional force.
The answer field takes the result of part b) in newtons.
Stopping distance of the block at different initial speeds (example data) $v_0\,(\text{m/s})$ $s\,(\text{m})$ 1.0 0.22 1.5 0.47 2.0 0.88 2.5 1.35 3.0 1.98 Answer
$F_\mu \approx 1.1\,\text{N}$
Show the solution
Given
$m = 0.50\,\text{kg}$, final speed $0$; initial speeds $v_0 = 1.0$; $1.5$; $2.0$; $2.5$; $3.0\,\text{m/s}$ and stopping distances $s = 0.22$; $0.47$; $0.88$; $1.35$; $1.98\,\text{m}$; the frictional force $F_\mu$ is constant, air resistance is ignored
Find
the frictional force $F_\mu$ in $\text{N}$
Principle
The table is horizontal, so the potential energy does not change. The weight and the normal force are perpendicular to the motion and do no work, so friction is the net force and its work equals the change in kinetic energy.
Reading the graph
The squares are $v_0^2 = 1.00$; $2.25$; $4.00$; $6.25$; $9.00\,\text{m}^2/\text{s}^2$. The points $(v_0^2, s)$ lie close to a straight line through the origin. The line passes through the point $(9.00\,\text{m}^2/\text{s}^2$; $1.97\,\text{m})$, so the slope is
$k$$= 1.97\,\text{m} : 9.00\,\text{m}^2/\text{s}^2$$= 0.219\,\text{s}^2/\text{m}$Formula
$-F_\mu s = 0 - \tfrac{1}{2} m v_0^2$, so $s = \dfrac{m}{2F_\mu} \cdot v_0^2$. The slope is $k = \dfrac{m}{2F_\mu}$, so $F_\mu = \dfrac{m}{2k}$.
Conversions
The quantities are in SI units, so no conversions are needed.
$1\,\text{kg} : (\text{s}^2/\text{m})$$= 1\,\text{kg}\cdot\text{m/s}^2$$= 1\,\text{N}$Substitution
$F_\mu$$= \dfrac{0.50\,\text{kg}}{2 \cdot 0.2188\,\text{s}^2/\text{m}}$$= 1.14\,\text{N}$Answer
$F_\mu \approx 1.1\,\text{N}$
Two significant figures like the mass $0.50\,\text{kg}$.
Reasonableness
The weight of the block is $0.50\,\text{kg} \cdot 9.81\,\text{m/s}^2 = 4.9\,\text{N}$, so the frictional force is about $23\,\%$ of the weight, that is the coefficient of friction is about $0.23$, which is believable for wood on a table. A single measurement gives the same: $v_0 = 2.0\,\text{m/s}$ gives
$F_\mu$$= m v_0^2 : (2s)$$= 0.50 \cdot 4.0 : 1.76\,\text{N}$$= 1.14\,\text{N}$
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 11-16
A cyclist coasts down a hill without pedalling or braking. The combined mass of the cyclist and the bicycle is $80\,\text{kg}$.
The speed is $4.0\,\text{m/s}$ at the top of the hill and $13.0\,\text{m/s}$ at the bottom. The height of the hill is $12.0\,\text{m}$, and the road down the hill is $150\,\text{m}$ long.
What is the average resistive force that slows the cyclist, when air resistance and rolling resistance are added together?
Answer
$22\,\text{N}$
Show the solution
Given
$m = 80\,\text{kg}$, speed at the top of the hill $v_1 = 4.0\,\text{m/s}$ and at the bottom $v_2 = 13.0\,\text{m/s}$, height $h = 12.0\,\text{m}$, length of the road $s = 150\,\text{m}$, $g = 9.81\,\text{m/s}^2$
Find
the average resistive force $F_\text{res}$ in $\text{N}$
Principle
The weight, the normal force of the road, and air resistance and rolling resistance act on the cyclist. The normal force does no work. The resistive forces are forces other than the weight, so the work–energy principle gives their work. The resistive force points against the motion all the way, so its work is $-F_\text{res} s$. The zero level is chosen at the bottom of the hill.
Formula
$\tfrac{1}{2} mv_1^2 + mgh - F_\text{res} s = \tfrac{1}{2} mv_2^2$, so
$F_\text{res}$$= \dfrac{\tfrac{m}{2}(v_1^2 - v_2^2) + mgh}{s}$Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
First
$\tfrac{m}{2}(v_1^2 - v_2^2)$$= \tfrac12 \cdot 80\,\text{kg} \cdot (16 - 169)\,\text{m}^2/\text{s}^2$$= -6120\,\text{J}$$mgh$$= 80\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 12.0\,\text{m}$$= 9417.6\,\text{J}$so
$F_\text{res}$$= \dfrac{-6120\,\text{J} + 9417.6\,\text{J}}{150\,\text{m}}$$= \dfrac{3297.6\,\text{J}}{150\,\text{m}}$$= 21.98\,\text{N}$Answer
$22\,\text{N}$
Two significant figures like the mass $80\,\text{kg}$ and the speed $4.0\,\text{m/s}$.
Reasonableness
The resistive forces took $3.3\,\text{kJ}$, about a third of the potential energy $9.4\,\text{kJ}$. At a speed of about $13\,\text{m/s}$ the air resistance on a cyclist is a few tens of newtons, so the average $22\,\text{N}$ for the whole hill is believable.
- Exercise 11-17
A roller coaster car with a mass of $500\,\text{kg}$ including passengers goes over the top of a $35.0\,\text{m}$ high hill at a speed of $2.0\,\text{m/s}$. The track to the top of the next hill is $120\,\text{m}$ long, and an average resistive force of $150\,\text{N}$ acts on the car all the way.
What is the greatest height the next hill may have so that the car reaches its top?
Answer
$31.5\,\text{m}$
Show the solution
Given
$m = 500\,\text{kg}$, height of the first top $h_1 = 35.0\,\text{m}$, speed there $v_1 = 2.0\,\text{m/s}$, track distance $s = 120\,\text{m}$, resistive force $F_\text{res} = 150\,\text{N}$, $g = 9.81\,\text{m/s}^2$
Find
the greatest height $h_2$ of the next hill in $\text{m}$
Principle
The weight, the normal force of the track and the resistive force act on the car. The normal force does no work, and the resistive force does the work $-F_\text{res} s$, so we use the work–energy principle. The car reaches the highest possible hill when its speed at the top is just zero. The zero level is chosen at the ground.
Formula
$\tfrac{1}{2} mv_1^2 + mgh_1 - F_\text{res} s = mgh_2$, so
$h_2$$= h_1 + \dfrac{v_1^2}{2 \cdot g} - \dfrac{F_\text{res} s}{mg}$Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$\dfrac{v_1^2}{2g}$$= \dfrac{(2.0\,\text{m/s})^2}{2 \cdot 9.81\,\text{m/s}^2}$$= 0.204\,\text{m}$$\dfrac{F_\text{res} s}{mg}$$= \dfrac{150\,\text{N} \cdot 120\,\text{m}}{500\,\text{kg} \cdot 9.81\,\text{m/s}^2}$$= 3.670\,\text{m}$so
$h_2$$= (35.0 + 0.204 - 3.670)\,\text{m}$$= 31.53\,\text{m}$Answer
at most $31.5\,\text{m}$, three significant figures like the height $35.0\,\text{m}$.
Reasonableness
The resistive force takes $150\,\text{N} \cdot 120\,\text{m} = 18\,\text{kJ}$, which corresponds to a height of about $3.7\,\text{m}$ for a $500\,\text{kg}$ car, and the small initial speed adds only $0.2\,\text{m}$. So the hill is a little lower than the previous one, as in roller coasters. In reality the hill is made even lower so that the car goes over the top at speed.
- Exercise 11-18
A stone is thrown at the seaside from a cliff $20.0\,\text{m}$ high at a speed of $12.0\,\text{m/s}$ obliquely upward at an angle of $35^\circ$ to the horizontal. Air resistance is ignored.
At what speed does the stone hit the water?
Answer
$23.2\,\text{m/s}$
Show the solution
Given
height $h = 20.0\,\text{m}$, initial speed $v_1 = 12.0\,\text{m/s}$ at an angle of $35^\circ$ to the horizontal, $g = 9.81\,\text{m/s}^2$; air resistance is ignored
Find
the speed $v_2$ of the stone at the water surface in $\text{m/s}$
Principle
After the throw only the weight acts on the stone, so the mechanical energy is conserved. Energy is a scalar: the kinetic energy $\tfrac{1}{2} mv^2$ depends only on the magnitude of the velocity, not on its direction, so the angle of the throw is not needed. The zero level is chosen at the water surface.
Formula
$\tfrac{1}{2} mv_1^2 + mgh = \tfrac{1}{2} mv_2^2$, so $v_2 = \sqrt{v_1^2 + 2gh}$
Conversions
The quantities are in SI units, so no conversions are needed. The mass cancels.
Substitution
First
$v_1^2$$= (12.0\,\text{m/s})^2$$= 144\,\text{m}^2/\text{s}^2$$2gh$$= 2 \cdot 9.81\,\text{m/s}^2 \cdot 20.0\,\text{m}$$= 392.4\,\text{m}^2/\text{s}^2$so
$v_2$$= \sqrt{(144 + 392.4)\,\text{m}^2/\text{s}^2}$$= 23.16\,\text{m/s}$Answer
$23.2\,\text{m/s}$
Three significant figures like the height $20.0\,\text{m}$ and the speed $12.0\,\text{m/s}$.
Reasonableness
The result is greater than the speed from a plain drop, $\sqrt{392.4}\,\text{m/s} \approx 19.8\,\text{m/s}$, and greater than the throwing speed, as it should be. If the stone were thrown with the same speed horizontally or downward, it would hit the water at the same speed $23.2\,\text{m/s}$ but in a different direction.
- Exercise 11-19
A snowboarder of mass $70\,\text{kg}$ including equipment goes down from the top of a slope, starting from rest. The height difference of the slope is $6.0\,\text{m}$ and the length of the slope $45\,\text{m}$. From the bottom of the slope the snowboarder glides on horizontal snow until stopping, and measures the distance $s$ travelled before stopping. The table is example data from six runs. Air resistance is ignored, and the frictional force of the snow is assumed to be constant both on the slope and on the horizontal part.
a) Calculate the mean stopping distance.
b) Calculate the average frictional force of the snow with the work–energy principle.
c) Calculate the speed of the snowboarder at the bottom of the slope and justify why it is smaller than in the frictionless case.
The answer field takes the result of part b).
Stopping distance on horizontal snow (example data) Run $s\,(\text{m})$ 1 29.4 2 31.2 3 28.7 4 30.5 5 29.9 6 30.7 Answer
$F \approx 55\,\text{N}$
Show the solution
Given
$m = 70\,\text{kg}$, height difference $h = 6.0\,\text{m}$, length of the slope $L = 45\,\text{m}$, initial speed $0$ and speed $0$ after stopping; horizontal stopping distance $s = 29.4$; $31.2$; $28.7$; $30.5$; $29.9$; $30.7\,\text{m}$, $g = 9.81\,\text{m/s}^2$; the frictional force $F$ is constant, air resistance is ignored
Find
a) the mean stopping distance $s_\text{mean}$; b) the average frictional force $F$ in $\text{N}$; c) the speed $v$ at the bottom of the slope
Principle
The weight, the normal force of the snow and friction act on the snowboarder. The normal force does no work, so friction is the only other force and we use the work–energy principle. Friction opposes the motion all the time, so its work is $-F$ times the whole distance travelled, $L + s_\text{mean}$. The zero level is the bottom of the slope. At the start and at the end the speed is zero, so all of the initial potential energy $mgh$ turns into internal energy.
Formula
a) $s_\text{mean} = \dfrac{\text{sum}}{6}$. b) $mgh - F\left(L + s_\text{mean}\right) = 0$, so $F = \dfrac{mgh}{L + s_\text{mean}}$. c) On the slope $mgh - F L = \tfrac{1}{2} mv^2$, so
$v$$= \sqrt{\dfrac{2\left(mgh - F L\right)}{m}}$Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
a) The sum of the stopping distances is $180.4\,\text{m}$, so
$s_\text{mean}$$= 180.4\,\text{m} : 6$$= 30.07\,\text{m}$b)
$mgh$$= 70\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 6.0\,\text{m}$$= 4120\,\text{J}$$L + s_\text{mean}$$= 45\,\text{m} + 30.07\,\text{m}$$= 75.07\,\text{m}$so
$F$$= 4120\,\text{J} : 75.07\,\text{m}$$= 54.9\,\text{N}$c) The work of friction on the slope is
$F L$$= 54.9\,\text{N} \cdot 45\,\text{m}$$= 2470\,\text{J}$so
$\tfrac{1}{2} mv^2$$= 4120\,\text{J} - 2470\,\text{J}$$= 1650\,\text{J}$$v$$= \sqrt{2 \cdot 1650\,\text{J} : 70\,\text{kg}}$$= 6.87\,\text{m/s}$Answer
b) $F \approx 55\,\text{N}$, two significant figures like the mass $70\,\text{kg}$ and the height $6.0\,\text{m}$. c) The speed at the bottom of the slope is about $6.9\,\text{m/s}$, and it is smaller than the speed on a frictionless slope, $\sqrt{2gh} = 10.8\,\text{m/s}$, because friction has already turned part of the potential energy into internal energy on the slope: the mechanical energy at the bottom of the slope is $1650\,\text{J}$ and not $4120\,\text{J}$.
Reasonableness
The frictional force $55\,\text{N}$ is about $8\,\%$ of the snowboarder's weight $687\,\text{N}$, that is the coefficient of friction is about $0.08$, which suits slippery snow. On the slope friction turns $2470\,\text{J} : 4120\,\text{J} \approx 60\,\%$ of the energy into internal energy, which corresponds to the share of the slope in the whole distance, $45 : 75.07 \approx 60\,\%$.
- Exercise 11-20
A tennis ball of mass $0.057\,\text{kg}$ is dropped from rest onto a hard floor from a height of $1.20\,\text{m}$. The height of the ball at its highest point after each bounce is measured from a video. The table is example data. Air resistance is ignored, so the mechanical energy changes only in the bounces. The zero level is the floor.
a) Calculate the mechanical energy of the ball at each highest point and the share that the ball keeps from each bounce.
b) Calculate the mean of the conserved shares in percent.
c) Calculate how much mechanical energy turns into internal energy in the first bounce. Justify with the data whether the mechanical energy of the ball is conserved in the bounces and whether the total energy is conserved.
The answer field takes the result of part b).
Height of the ball at the highest point (example data) Bounces $h\,(\text{m})$ 0 1.20 1 0.77 2 0.49 3 0.32 4 0.20 Answer
$64\,\%$
Show the solution
Given
$m = 0.057\,\text{kg}$, $g = 9.81\,\text{m/s}^2$; heights at the top $h_0 = 1.20\,\text{m}$ and after the bounces $h_1 = 0.77$; $h_2 = 0.49$; $h_3 = 0.32$; $h_4 = 0.20\,\text{m}$; air resistance is ignored
Find
a) the mechanical energy of the ball at the top points and the share conserved in each bounce; b) the mean of the shares in percent; c) the mechanical energy turned into internal energy in the first bounce, and a justification about the conservation of energy
Principle
At the top the ball is momentarily at rest, so the kinetic energy is zero and $E_\text{mech} = mgh$ (the floor as the zero level). During the flight only the weight acts, so the mechanical energy is conserved, and the changes happen in the bounces. The mass and $g$ cancel from the share, so the share is the ratio of the heights, $\dfrac{E_{n+1}}{E_n} = \dfrac{h_{n+1}}{h_n}$.
Formula
$E_n = mgh_n$, the share $r_n = \dfrac{h_{n+1}}{h_n}$, and the energy change in the first bounce
$\Delta E$$= E_0 - E_1$$= mg\left(h_0 - h_1\right)$Conversions
The quantities are in SI units. The share is changed to percent by multiplying by one hundred.
Substitution
a) The energies are
$E_0$$= 0.057 \cdot 9.81 \cdot 1.20\,\text{J}$$= 0.671\,\text{J}$$E_1 = 0.431\,\text{J}$, $E_2 = 0.274\,\text{J}$, $E_3 = 0.179\,\text{J}$ and $E_4 = 0.112\,\text{J}$. The shares are $r_0 = 0.77 : 1.20 = 0.642$, $r_1 = 0.636$, $r_2 = 0.653$ and $r_3 = 0.625$. b) The sum of the shares is $2.556$, so the mean is $2.556 : 4 = 0.639 = 63.9\,\%$. c)
$\Delta E$$= 0.057\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot (1.20 - 0.77)\,\text{m}$$= 0.240\,\text{J}$Answer
b) In a bounce the ball keeps on average about $64\,\%$ of its mechanical energy, two significant figures like the heights. c) In the first bounce about $0.24\,\text{J}$ of mechanical energy turns into internal energy. Justification: the mechanical energy of the ball is not conserved in the bounces, because it decreases every time (the heights decrease: $1.20 \to 0.77 \to 0.49 \to 0.32 \to 0.20\,\text{m}$), but the total energy is conserved: in a bounce the interaction between the ball and the floor turns part of the mechanical energy into internal energy (warming) of the ball and the floor, and into sound.
Reasonableness
The shares $0.625$–$0.653$ are almost the same, so the ball loses about the same share in every bounce, which is believable for a tennis ball. After four bounces $0.20 : 1.20 \approx 17\,\%$ of the initial energy is left, and $0.639^4 \approx 0.167$ gives the same.