Newton's second law
Why does a fully loaded truck accelerate away from the traffic lights more slowly than an empty one, even though the engine is the same?
Prior knowledge
- Acceleration is the change in velocity per unit of time, $a = \dfrac{\Delta v}{\Delta t}$chapter Acceleration and uniformly accelerated motion
- When the forces acting on an object cancel each other, the object is at rest or moves at constant velocitychapter Interaction and force: Newton's first and third laws
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 4.
Think about it
Two identical timber trucks stand side by side at the traffic lights. One is empty, the other is full of logs. The light turns green, and both drivers accelerate with equal effort. Within a few seconds the empty truck is clearly ahead.
What decides how quickly the velocity of an object changes, and why does the same engine not get the loaded truck up to speed as quickly?
Optional Recall: uniform motion
Predict
Make your predictions before investigating 3 questions
Prediction 1: the force doubles
Prediction 2: the mass doubles
Optional Prediction 3: a force against the motion
Investigate
In the simulation a string pulls a laboratory cart on a horizontal track. At the top is the situation picture, on the right the free-body diagram of the cart and at the bottom the velocity graph, whose slope is the acceleration.
Simulation
How to use
Adjust the pulling force $F$ and the mass $m$. Choose whether the cart is at rest or moving to the right at the start, and which way the string pulls. Press Start. The run lasts $1.0\,\text{s}$ or ends before the end of the track. After the run an orange straight line is drawn in the graph, and its slope $a_x$ is shown in the readouts. Reset returns the cart to the start. Speed slows down or speeds up the motion without changing the readouts.
Assumptions of the model
The values are model values calculated by the simulation. The track is horizontal, and friction and air resistance are not taken into account. The weight $\overline{G}$, the normal force $\overline{N}$ from the track and the force $\overline{F}$ of the string act on the cart. The weight and the normal force cancel each other, so the net force is the force of the string. The force of the string stays constant throughout the run. The arrows in the free-body diagram are on one scale, so the pulling force looks short next to the weight.
How the calculation works
The positive direction is to the right. The velocity $v_x$ of the cart is recorded in the graph every $0.05\,\text{s}$. The acceleration $a_x$ is the slope $\dfrac{\Delta v_x}{\Delta t}$ of the straight line fitted to the points. In the readouts $\Sigma F_x$ is the force of the string with its sign and $G = mg$, where $g = 9.81\,\text{m/s}^2$.
Experiment 1: the pulling force changes
Test the first prediction. The starting state selects a mass of $1.0\,\text{kg}$ and a cart at rest, and the string pulls to the right. Change only the force $F$. Press Start each time and record the measurement when the run has ended. Try at least the forces $1.0\,\text{N}$ and $2.0\,\text{N}$.
Initial state: initial state: at rest, string pulls: to the right, mass $m$: $1.0\,\text{kg}$, force $F$ = $1\,\text{N}$. Kept constant: initial state, string pulls, mass $m$. Varied: force $F$.
| force $F$ | $\Sigma F_x$ $(\text{N})$ | $a_x$ (m/s$^2$) |
|---|
Experiment 2: the mass changes
Test the second prediction. The starting state selects a force of $1.0\,\text{N}$ and a cart at rest, and the string pulls to the right. Change only the mass $m$. Press Start and record a measurement for each mass.
Initial state: initial state: at rest, string pulls: to the right, force $F$ = $1\,\text{N}$, mass $m$: $0.50\,\text{kg}$. Kept constant: initial state, string pulls, force $F$. Varied: mass $m$.
| mass $m$ | $m$ $(\text{kg})$ | $a_x$ (m/s$^2$) |
|---|
Optional Experiment 3: a force against the motion
Test the third prediction. The starting state selects a force of $2.0\,\text{N}$, a mass of $1.0\,\text{kg}$ and a cart that moves to the right at the start. Change only the direction of the string. During the run, watch the blue and the orange arrow and the points of the graph. Record a measurement for both directions.
Initial state: initial state: moving to the right, string pulls: to the right, force $F$ = $2\,\text{N}$, mass $m$: $1.0\,\text{kg}$. Kept constant: initial state, force $F$, mass $m$. Varied: string pulls.
| string pulls | $\Sigma F_x$ $(\text{N})$ | $a_x$ (m/s$^2$) | $v_x$ at the end $(\text{m/s})$ |
|---|
Conclusion
Using your measurements, explain how the acceleration of the cart depends on the net force and the mass. Then explain why a loaded truck accelerates more slowly than an empty one when the force of the road on the tyres is the same.
Model answer
In experiment 1 the mass was $1.0\,\text{kg}$. In the values of the model the acceleration was $1.0\,\text{m/s}^2$ with a force of $1.0\,\text{N}$ and $2.0\,\text{m/s}^2$ with a force of $2.0\,\text{N}$: a doubled net force gave a doubled acceleration. In experiment 2 the force was $1.0\,\text{N}$. The acceleration was $2.0\,\text{m/s}^2$ with a mass of $0.50\,\text{kg}$ and $1.0\,\text{m/s}^2$ with a mass of $1.0\,\text{kg}$: a doubled mass halved the acceleration. So the acceleration is proportional to the net force and inversely proportional to the mass, $a = \dfrac{\Sigma F}{m}$. The mass of a loaded truck is many times that of an empty one. When the force of the road on the tyres is the same, the acceleration is that many times smaller. The weight of the load does not pull the truck backward, because the normal force from the road cancels it.
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
The net force changes the velocity
In the chapter Interaction and force: Newton's first and third laws you learned that an object's velocity stays unchanged when the forces acting on it cancel each other. The object is then at rest or moves in a straight line at constant velocity. When the forces do not cancel, the velocity changes: the object speeds up, slows down or turns. What matters is not any single force but the combined effect of all the forces acting on the object, the net force $\Sigma\overline{F}$. It is the vector sum of the forces, not just one force added to another.
On the left of the figure a box rests on the floor. The weight $\overline{G}$ and the normal force $\overline{N}$ from the floor are equal and opposite, so the net force is zero. On the right the box is pulled by a rope. The vertical forces still cancel, but horizontally the tension $\overline{T}$ in the rope is greater than the friction force $\overline{F}_\mu$ from the floor. The net force points to the right, and the box accelerates to the right.
Mass resists a change in velocity
Measurements with a cart reveal two regularities. When the net force doubles, the acceleration doubles: the acceleration is directly proportional to the net force. When the same net force acts on a cart with twice the mass, the acceleration is halved. Mass therefore tells us about the inertia of an object, that is, how hard it is to change its velocity. In the graph each mass gives a straight line through the origin, and the line of the heavier cart is half as steep.
This gives the answer to the opening question. The engine turns the driving wheels, and the road pushes the tyres forward. When this force is equally large for the loaded and the empty truck, the much greater mass of the loaded truck makes its acceleration small. The weight of the load does not pull the truck backward as such, because the normal force from the road cancels it. The heavy load does increase the rolling resistance of the tyres, but the main reason for the slow acceleration is the large mass.
Newton's second law
Newton's second law, also called the fundamental law of dynamics, gathers the observations into one equation:
$$\Sigma\overline{F} = m\overline{a}$$
where $\Sigma\overline{F}$ is the vector sum of the forces acting on the object, the net force ($\text{N}$), $m$ is the mass of the object ($\text{kg}$) and $\overline{a}$ is its acceleration ($\text{m/s}^2$). The unit of force is defined by this law: one newton is the force that gives a mass of one kilogram an acceleration of $1\,\text{m/s}^2$. So $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$.
The net force points in the direction of the acceleration
The law is a vector equation: the net force and the acceleration always point in the same direction. The direction of the velocity can be another one. In the figure a crane holds a crate on a rope. When the crate rises while speeding up, the force of the rope is greater than the weight and the net force points up. When the crate rises while slowing down, the force of the rope is smaller than the weight: the net force and the acceleration point down, even though the crate still moves upward. At rest and when rising at constant velocity, the force of the rope and the weight are equal.
First the forces, then the equation
When the law is used in a calculation, the steps are always taken in the same order. Choose the object and draw its free-body diagram, in which every force has a cause: the Earth, a rope, a surface or a puller. Choose a positive direction, usually the direction of the acceleration. Only then write the net force from the forces, with signs, and set it equal to $ma$. For the crate of the crane, when up is positive, we get $T - G = ma$. In the chapter Free-body diagrams and the combined effect of forces the same method extends to slanted forces.
Common misconception
Many people think that a force always acts in the direction of motion and that the net force points in the direction of motion. The net force, however, points in the direction of the acceleration. When air resistance is ignored, a ball thrown upward keeps rising even though only its weight, which points down, acts on it, and that is why it slows down. In a slowing motion the net force points against the velocity. A second mistake is to think that a constant force keeps up a constant velocity. When the net force is constant and not zero, the velocity changes steadily. A third mistake is to confuse mass and weight. On a horizontal track the weight does not resist the motion, because the normal force cancels it: it is the mass that makes the acceleration smaller.
Summary
- When the net force is zero, the velocity does not change. When it is not zero, the object accelerates in the direction of the net force.
- Newton's second law is $\Sigma\overline{F} = m\overline{a}$. The acceleration is directly proportional to the net force and inversely proportional to the mass.
- The net force points in the direction of the acceleration, which is not always the direction of the velocity: in a slowing motion it points against the velocity.
- First the free-body diagram is drawn and a positive direction is chosen, then $\Sigma F = ma$ is written with signs.
- The unit of force is $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$.
Optional Return to your predictions
Compare your predictions with the measurements. What surprised you most? Write in your own words what the direction of the net force tells you and what it does not tell you.
Would you change your prediction?
Apply
Worked example
Problem
A construction crane lifts a concrete slab of mass $480\,\text{kg}$. The tension in the rope is $5.20\,\text{kN}$. What is the acceleration of the slab, and in which direction does it point? Air resistance is ignored.
Step 1/8
Given and find. $m = 480\,\text{kg}$, $T = 5.20\,\text{kN}$ and $g = 9.81\,\text{m/s}^2$. We want the acceleration $a$ and its direction.
Step 2/8
Principle. Choose the concrete slab as the object. The Earth (weight) and the rope (tension) act on it. By Newton's second law, the acceleration is determined by the sum of these forces. The force of the rope alone does not accelerate the slab, because most of it is used to hold up the weight.
Step 3/8
Free-body diagram. The slab is drawn as a rectangle. The weight $\overline{G}$ starts from the centre and points down, and the tension $\overline{T}$ starts from the top surface and points up. The positive direction is chosen to be up.
Step 4/8
Rearranged formula. From the diagram we write the net force $T - mg$ and solve for the acceleration:
$$\begin{aligned} &T - mg = ma \\ &a = \frac{T - mg}{m} \end{aligned}$$
Step 5/8
Unit conversions. The tension is converted to newtons:
Step 6/8
Substitution. First we calculate the weight and then the acceleration:
$$\begin{aligned} a &= \frac{T - mg}{m} \\ mg &= 480\,\text{kg}\cdot 9.81\,\tfrac{\text{m}}{\text{s}^2} \\ \phantom{mg} &= 4708.8\,\text{N} \\ a &= \frac{5200\,\text{N} - 4708.8\,\text{N}}{480\,\text{kg}} \\ \phantom{a} &\approx 1.02\,\text{m/s}^2 \end{aligned}$$
Step 7/8
Result. The acceleration is $a \approx 1.02\,\text{m/s}^2$ upward. It has three significant figures, like the tension $5.20\,\text{kN}$ and the mass $480\,\text{kg}$.
Step 8/8
Reasonableness. The force of the rope is only about ten percent greater than the weight $4.71\,\text{kN}$, so the acceleration should also be about a tenth of the free-fall acceleration: $0.1\cdot 9.81\,\text{m/s}^2 \approx 1\,\text{m/s}^2$. If the weight were forgotten, the result would be $5200\,\text{N} : 480\,\text{kg} \approx 10.8\,\text{m/s}^2$. That is more than the free-fall acceleration and far too large for a crane.
Choose the correct free-body diagram
Apply to a new situation
A passenger car of mass $1400\,\text{kg}$ is driving on a horizontal road at a velocity of $72\,\text{km/h}$. The driver brakes, and the car stops, slowing down uniformly, in $4.0\,\text{s}$.
What is the net force acting on the car during the braking? Give the answer in newtons.
Hint 1: a guiding question
The net force is obtained from the law $\Sigma F = ma$. What do you need to know first, and where do you get it?
Hint 2: what is the connection
The acceleration is the change in velocity divided by the time: $a = \dfrac{\Delta v}{\Delta t}$. First convert the velocity to $\text{m/s}$.
Hint 3: an intermediate step
$72\,\text{km/h} = 20\,\text{m/s}$, so
Multiply by the mass.
Show the solution
Given
$m = 1400\,\text{kg}$, initial velocity $v_0 = 72\,\text{km/h}$, final velocity $v = 0$, time $\Delta t = 4.0\,\text{s}$, the car slows down uniformly
Find
the magnitude of the net force $\Sigma F$ in $\text{N}$
Principle
The velocity of the car changes steadily, so its acceleration is constant. By Newton's second law the net force is the mass times the acceleration, and it points in the direction of the acceleration, that is backward.
Free-body diagram
The car as a rectangle, the positive direction the direction of motion. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ from the road points up from the tyres, equally long. The friction force $\overline{F}_\mu$ from the road points backward from the contact point of the tyres. There is no force pushing forward.
Formula
$a = \dfrac{v - v_0}{\Delta t}$ and $\Sigma F = ma$, so
Conversions
Substitution
Answer
$7000\,\text{N}$
Or $7.0\,\text{kN}$ backward, two significant figures like the time $4.0\,\text{s}$ and the velocity $72\,\text{km/h}$. The minus sign tells us that the force points against the motion.
Reasonableness
The weight of the car is about $14\,\text{kN}$, so the braking force is about half of the weight. The deceleration $5.0\,\text{m/s}^2$ is about half of the free-fall acceleration, which is an ordinary firm braking on a dry road. The force points backward even though the car moves forward, because the car is slowing down.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- The net force acting on an object doubles, and at the same time the mass of the object is halved. By what factor does the acceleration change?
Answer
It becomes four times as great. The acceleration $a = \dfrac{\Sigma F}{m}$ doubles because of the force and doubles once more when the mass is halved: $2\cdot 2 = 4$.
- An elevator moves upward and slows down as it stops at a floor. Which is greater during the slowing down, the tension in the supporting rope or the weight of the elevator? Justify your answer.
Answer
The weight is greater. The elevator slows down while moving up, so its acceleration and net force point down: $T - G = ma$, where $a < 0$, so $T < G$.
- What is one newton? Express it using kilograms, metres and seconds.
Answer
One newton is the force that gives a mass of $1\,\text{kg}$ an acceleration of $1\,\text{m/s}^2$. So $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$.
1. Concept
2. Calculation
A shopping trolley of mass $18\,\text{kg}$ is pushed on a horizontal floor with a force of $30\,\text{N}$. The rolling resistance of the wheels and other friction slow the trolley with a total force of $12\,\text{N}$.
What is the acceleration of the trolley? Give the answer in $\text{m/s}^2$.
Hint 1: a guiding question
Draw the free-body diagram of the trolley. Which forces are horizontal, and which cancel each other in the vertical direction?
Hint 2: what is the connection
Choose the direction of the push as the positive direction. In the horizontal direction $F - F_\mu = ma$.
Hint 3: an intermediate step
Show the solution
Given
$m = 18\,\text{kg}$, pushing force $F = 30\,\text{N}$ horizontally, slowing forces in total $F_\mu = 12\,\text{N}$
Find
the acceleration $a$ of the trolley in $\text{m/s}^2$
Principle
By Newton's second law the acceleration of the trolley is determined by the net force. In the vertical direction the trolley does not accelerate, so only the horizontal forces remain in the net force.
Free-body diagram
The trolley as a rectangle, the positive direction the direction of the push. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ from the floor points up from the bottom surface, equally long. The pushing force $\overline{F}$ points forward and the slowing force $\overline{F}_\mu$ points backward from the edge of the bottom surface, shorter than the pushing force.
Formula
$F - F_\mu = ma$, so $a = \dfrac{F - F_\mu}{m}$
Conversions
All quantities are in SI units, and $\tfrac{\text{N}}{\text{kg}} = \tfrac{\text{m}}{\text{s}^2}$, so no conversions are needed.
Substitution
Answer
$1.0\,\text{m/s}^2$
In the direction of the push, two significant figures like the data $30\,\text{N}$, $12\,\text{N}$ and $18\,\text{kg}$.
Reasonableness
Without the slowing forces the acceleration would be $30\,\text{N} : 18\,\text{kg} \approx 1.7\,\text{m/s}^2$, so friction makes it smaller, as it should. An acceleration of $1\,\text{m/s}^2$ means that the velocity of the trolley increases by $1\,\text{m/s}$ every second, which fits a brisk push.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Draw it yourself
A book of mass $1.2\,\text{kg}$ lies on a horizontal table. You push it horizontally with your hand with a force of $5.0\,\text{N}$, and the friction force from the table on the book is $3.0\,\text{N}$. Draw the free-body diagram of the book on paper: the book as a rectangle and every force from its point of action. Mark the direction of the acceleration separately next to the diagram. Then write the equations of motion in the horizontal and vertical directions and calculate the acceleration of the book.
Model diagram
The Earth (weight $\overline{G}$), the table (normal force $\overline{N}$ and friction force $\overline{F}_\mu$) and the hand (pushing force $\overline{F}$) act on the book. In the vertical direction the book does not accelerate, so $N - mg = 0$ and $N = mg = 1.2\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 11.8\,\text{N}$. In the horizontal direction, when the positive direction is the direction of the push, $F - F_\mu = ma$, so $a = \dfrac{F - F_\mu}{m} = \dfrac{5.0\,\text{N} - 3.0\,\text{N}}{1.2\,\text{kg}} \approx 1.7\,\text{m/s}^2$. The net force $2.0\,\text{N}$ and the acceleration point in the direction of the push.
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.
- Newton's second law
- A net force gives an object an acceleration; the fundamental law of dynamics.$$\Sigma\overline{F} = m\overline{a}$$
- Net force
- The vector sum of the forces acting on an object.
Exercises
Level 1 · basic practice 8 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 5-1Core
A horizontal net force of $6.0\,\text{N}$ acts on a toy car of mass $2.5\,\text{kg}$. What is the acceleration of the toy car?
Answer
$2.4\,\text{m/s}^2$
Show the solution
Given
$m = 2.5\,\text{kg}$, net force $\Sigma F = 6.0\,\text{N}$ horizontally
Find
the acceleration $a$ in $\text{m/s}^2$
Principle
By Newton's second law the net force determines the acceleration: $\Sigma F = ma$. The net force is given directly, so the forces do not need to be added.
Formula
$\Sigma F = ma$, so $a = \dfrac{\Sigma F}{m}$
Conversions
The quantities are in SI units, and $\tfrac{\text{N}}{\text{kg}} = \tfrac{\text{m}}{\text{s}^2}$, so no conversions are needed.
Substitution
$a$$= \dfrac{6.0\,\text{N}}{2.5\,\text{kg}}$$= 2.4\,\text{m/s}^2$Answer
$2.4\,\text{m/s}^2$
In the direction of the net force, two significant figures like the data $6.0\,\text{N}$ and $2.5\,\text{kg}$.
Reasonableness
The net force has a little more than twice as many newtons as the mass has kilograms, so the acceleration is a little over $2\,\text{m/s}^2$. The speed of the toy car increases by $2.4\,\text{m/s}$ every second, which fits a brisk push.
- Exercise 5-2Core
A cyclist and a bicycle with a total mass of $80\,\text{kg}$ accelerate on a horizontal road with an acceleration of $1.5\,\text{m/s}^2$.
What is the net force acting on them?
Answer
$120\,\text{N}$
Show the solution
Given
total mass $m = 80\,\text{kg}$, acceleration $a = 1.5\,\text{m/s}^2$
Find
the net force $\Sigma F$ in $\text{N}$
Principle
Newton's second law: the acceleration is produced by the net force, and $\Sigma F = ma$. The cyclist and the bicycle are treated as one object because they move together.
Formula
$\Sigma F = ma$
Conversions
The quantities are in SI units, and $\text{kg}\cdot\tfrac{\text{m}}{\text{s}^2} = \text{N}$, so no conversions are needed.
Substitution
$\Sigma F$$= 80\,\text{kg}\cdot 1.5\,\text{m/s}^2$$= 120\,\text{N}$Answer
$120\,\text{N}$
Or $1.2\cdot 10^{2}\,\text{N}$ in the direction of motion, two significant figures like the mass $80\,\text{kg}$ and the acceleration $1.5\,\text{m/s}^2$.
Reasonableness
The net force is about one sixth of the weight of the cyclist and the bicycle, $80\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 785\,\text{N}$, so the acceleration is about one sixth of the free-fall acceleration, $1.5\,\text{m/s}^2$. The net force is what is left when the resistive forces are subtracted from the force of the road on the tyre.
- Exercise 5-3
A net force of $45\,\text{N}$ acts on an object, and its acceleration is $3.0\,\text{m/s}^2$. What is the mass of the object?
Answer
$15\,\text{kg}$
Show the solution
Given
net force $\Sigma F = 45\,\text{N}$, acceleration $a = 3.0\,\text{m/s}^2$
Find
the mass $m$ in $\text{kg}$
Principle
By Newton's second law $\Sigma F = ma$. When the net force and the acceleration are known, the mass is obtained by solving the law for the mass: the mass tells how much force is needed for a given acceleration.
Formula
$\Sigma F = ma$, so $m = \dfrac{\Sigma F}{a}$
Conversions
The quantities are in SI units, and
$\dfrac{\text{N}}{\text{m/s}^2}$$= \dfrac{\text{kg}\cdot\text{m/s}^2}{\text{m/s}^2}$$= \text{kg}$so no conversions are needed.
Substitution
$m$$= \dfrac{45\,\text{N}}{3.0\,\text{m/s}^2}$$= 15\,\text{kg}$Answer
$15\,\text{kg}$
Two significant figures like the data $45\,\text{N}$ and $3.0\,\text{m/s}^2$.
Reasonableness
Check by multiplying: $15\,\text{kg}\cdot 3.0\,\text{m/s}^2 = 45\,\text{N}$, which is the given net force. The unit comes out as kilograms, as it must for a mass.
- Exercise 5-4
A box of mass $12\,\text{kg}$ is pulled on a horizontal floor with a horizontal force of $38\,\text{N}$. The friction force from the floor on the box is $20\,\text{N}$. Draw the free-body diagram of the box.
What is the acceleration of the box?
Answer
$1.5\,\text{m/s}^2$
Show the solution
Given
$m = 12\,\text{kg}$, pulling force $F = 38\,\text{N}$ horizontally, friction force $F_\mu = 20\,\text{N}$
Find
the acceleration $a$ of the box in $\text{m/s}^2$
Principle
By Newton's second law the acceleration is determined by the net force, not by the pulling force alone. In the vertical direction the box does not accelerate, so the vertical forces cancel and the net force is horizontal.
Free-body diagram
The box as a rectangle, the positive direction the direction of the pull. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ from the floor points up from the bottom surface, equally long. The pulling force $\overline{F}$ points forward and the friction force $\overline{F}_\mu$ from the floor points backward from the back corner of the bottom surface, shorter than the pulling force.
Formula
In the horizontal direction $F - F_\mu = ma$, so $a = \dfrac{F - F_\mu}{m}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$a$$= \dfrac{38\,\text{N} - 20\,\text{N}}{12\,\text{kg}}$$= \dfrac{18\,\text{N}}{12\,\text{kg}}$$= 1.5\,\text{m/s}^2$Answer
$1.5\,\text{m/s}^2$
In the direction of the pull, two significant figures like the data.
Reasonableness
Without friction the acceleration would be $38\,\text{N} : 12\,\text{kg} \approx 3.2\,\text{m/s}^2$. Friction takes more than half of the pulling force, so the acceleration is less than half of that, as it should be.
- Exercise 5-5
At a building site a bucket of mass $8.0\,\text{kg}$ is lifted straight up on a rope at constant velocity. Draw the free-body diagram of the bucket.
What is the tension in the rope?
Answer
$78\,\text{N}$
Show the solution
Given
$m = 8.0\,\text{kg}$, the bucket rises at constant velocity; $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ in the rope in $\text{N}$
Principle
In uniform motion the acceleration is zero, so by Newton's second law the net force is zero even though the bucket is moving.
Free-body diagram
The bucket as a rectangle, the positive direction up. The tension $\overline{T}$ of the rope points up from the top surface and the weight $\overline{G}$ points down from the centre, equally long. The velocity is not drawn as a force.
Formula
$T - mg = 0$, so $T = mg$
Conversions
The mass is in kilograms, so no conversions are needed.
Substitution
$T$$= 8.0\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 78.48\,\text{N}$Answer
$78\,\text{N}$
Two significant figures like the mass $8.0\,\text{kg}$.
Reasonableness
The weight of one kilogram is about $10\,\text{N}$, so the weight of an eight-kilogram bucket is about $80\,\text{N}$. The rope supports the whole weight and no more, because moving upward at constant velocity does not require a net force.
- Exercise 5-6
An elevator moves upward and slows down to stop. The positive direction is up. Which statement about the net force $\Sigma\overline{F}$ acting on the elevator is correct?
Answer
The net force points down, because the speed decreases and the acceleration points downward.
Show the solution
Find
in which direction the net force $\Sigma\overline{F}$ acting on the elevator points when the elevator rises and slows down
Principle
By Newton's second law $\Sigma\overline{F} = m\overline{a}$, so the net force always points in the direction of the acceleration. The direction of the acceleration is the direction of the change of velocity: when the speed decreases, the acceleration points against the velocity.
Model answer
The correct statement is the one saying that the net force points down, because the speed decreases and the acceleration points down. The elevator is still moving upward, but its speed decreases, so the acceleration and the net force point down. The force of the rope is then smaller than the weight. The statement that the direction of the velocity decides the direction of the net force confuses velocity and acceleration. The statement about a zero net force would hold only in a steady rise. The statement that the force of the rope is always greater than the weight holds only when the rise begins.
- Exercise 5-7
The mass of an ice hockey puck is $170\,\text{g}$. A hit from the stick gives the puck a horizontal acceleration of $400\,\text{m/s}^2$.
What is the net force acting on the puck during the hit?
Answer
$68\,\text{N}$
Show the solution
Given
$m = 170\,\text{g}$, $a = 400\,\text{m/s}^2$
Find
the net force $\Sigma F$ in $\text{N}$
Principle
By Newton's second law the acceleration is produced by the net force: $\Sigma F = ma$. The acceleration and the mass of the puck are known, so the net force is obtained once the mass is first in kilograms.
Formula
$\Sigma F = ma$
Conversions
$170\,\text{g}$$= 170 : 1000\,\text{kg}$$= 0.17\,\text{kg}$Substitution
$\Sigma F$$= 0.17\,\text{kg}\cdot 400\,\text{m/s}^2$$= 68\,\text{N}$Answer
$68\,\text{N}$
In the direction of motion, two significant figures like the mass $170\,\text{g}$.
Reasonableness
The weight of the puck is $0.17\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 1.7\,\text{N}$, so the hit is about $40$ times the weight, since the acceleration is about $40$ times the free-fall acceleration: $400 : 9.81 \approx 41$. A small puck needs a large force because the acceleration is large.
- Exercise 5-8
A horizontal net force of $0.60\,\text{kN}$ acts on a snowmobile, and its acceleration is $1.5\,\text{m/s}^2$. What is the mass of the snowmobile?
Answer
$4.0\cdot 10^{2}\,\text{kg}$
Show the solution
Given
$\Sigma F = 0.60\,\text{kN}$, $a = 1.5\,\text{m/s}^2$
Find
the mass $m$ in $\text{kg}$
Principle
Newton's second law $\Sigma F = ma$ connects the net force, the mass and the acceleration. When two of them are known, the third is obtained by solving the law for it.
Formula
$\Sigma F = ma$, so $m = \dfrac{\Sigma F}{a}$
Conversions
$0.60\,\text{kN}$$= 0.60 \cdot 1000\,\text{N}$$= 600\,\text{N}$Substitution
$m$$= \dfrac{600\,\text{N}}{1.5\,\text{m/s}^2}$$= 400\,\text{kg}$Answer
$4.0\cdot 10^{2}\,\text{kg}$
Or $400\,\text{kg}$, two significant figures like the data.
Reasonableness
Check by multiplying: $400\,\text{kg}\cdot 1.5\,\text{m/s}^2 = 600\,\text{N}$. The mass of a snowmobile is realistically a few hundred kilograms.
Level 2 · applied practice 7 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 5-9Core
A passenger car of mass $1200\,\text{kg}$ accelerates uniformly on a horizontal road from rest to a velocity of $90\,\text{km/h}$ in $10\,\text{s}$.
What is the net force acting on the car during the acceleration?
Answer
$3000\,\text{N}$
Show the solution
Given
$m = 1200\,\text{kg}$, initial velocity $v_0 = 0$, final velocity $v = 90\,\text{km/h}$, time $\Delta t = 10\,\text{s}$, uniform acceleration
Find
the net force $\Sigma F$ in $\text{N}$
Principle
The velocity of the car increases steadily, so its acceleration is constant and is obtained from the change in velocity. By Newton's second law the net force is the mass times the acceleration.
Formula
$a = \dfrac{v - v_0}{\Delta t}$ and $\Sigma F = ma$, so
$\Sigma F$$= \dfrac{m(v - v_0)}{\Delta t}$Conversions
$90\,\text{km/h}$$= 90 : 3.6\,\text{m/s}$$= 25\,\text{m/s}$Substitution
$\Sigma F$$= \dfrac{1200\,\text{kg}\cdot(25\,\text{m/s} - 0)}{10\,\text{s}}$$= 1200\,\text{kg}\cdot 2.5\,\text{m/s}^2$$= 3000\,\text{N}$Answer
$3000\,\text{N}$
Or $3.0\,\text{kN}$ in the direction of motion, two significant figures like the time $10\,\text{s}$ and the velocity $90\,\text{km/h}$.
Reasonableness
An acceleration of $2.5\,\text{m/s}^2$ is a brisk acceleration for an ordinary passenger car. The net force is about a quarter of the weight of the car, $1200\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 12\,\text{kN}$, just as the acceleration is about a quarter of the free-fall acceleration.
- Exercise 5-10Core
An elevator car with its passengers has a mass of $900\,\text{kg}$. It starts to move downward with an acceleration of $1.5\,\text{m/s}^2$.
What is the tension in the supporting rope during the acceleration?
Answer
$7500\,\text{N}$
Show the solution
Given
$m = 900\,\text{kg}$, downward acceleration $a = 1.5\,\text{m/s}^2$; $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ in the supporting rope in $\text{N}$
Principle
The elevator accelerates downward, so by Newton's second law the net force points down and equals $ma$. The weight is therefore greater than the force of the rope.
Free-body diagram
The elevator car with its passengers as a rectangle, the positive direction down, the direction of the acceleration. The weight $\overline{G}$ points down from the centre and the tension $\overline{T}$ of the rope points up from the top surface, shorter than the weight. The acceleration is marked separately, not as a force.
Formula
$mg - T = ma$, so $T = m(g - a)$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$T$$= 900\,\text{kg}\cdot(9.81\,\text{m/s}^2 - 1.5\,\text{m/s}^2)$$= 900\,\text{kg}\cdot 8.31\,\text{m/s}^2$$= 7479\,\text{N}$Answer
$7500\,\text{N}$
Or $7.5\,\text{kN}$, two significant figures like the acceleration $1.5\,\text{m/s}^2$.
Reasonableness
The weight of the elevator is $900\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 8.8\,\text{kN}$. The force of the rope is $ma = 1.35\,\text{kN}$ smaller than this, so that the net force points down. When an elevator starts to go down, a passenger feels lighter for a moment.
- Exercise 5-11
A crane lowers a load of $120\,\text{kg}$ downward at a velocity of $2.0\,\text{m/s}$. Just before the ground the load slows down uniformly and stops in $2.0\,\text{s}$.
What is the tension in the rope during the slowing down?
Answer
$1300\,\text{N}$
Show the solution
Given
$m = 120\,\text{kg}$, downward velocity $v_0 = 2.0\,\text{m/s}$, final velocity $v = 0$, time $\Delta t = 2.0\,\text{s}$, uniform slowing down; $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ in the rope in $\text{N}$
Principle
The load moves down but slows down, so its acceleration points up. By Newton's second law the net force points in the direction of the acceleration, that is up, so the force of the rope is greater than the weight.
Free-body diagram
The load as a rectangle, the positive direction up. The tension $\overline{T}$ of the rope points up from the top surface, longer than the weight, and the weight $\overline{G}$ points down from the centre. The downward velocity and the upward acceleration are marked separately next to the diagram.
Formula
The magnitude of the acceleration is $a = \dfrac{|v - v_0|}{\Delta t}$ upward, and $T - mg = ma$, so $T = m(g + a)$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$a$$= \dfrac{2.0\,\text{m/s}}{2.0\,\text{s}}$$= 1.0\,\text{m/s}^2$$T$$= 120\,\text{kg}\cdot(9.81\,\text{m/s}^2 + 1.0\,\text{m/s}^2)$$= 120\,\text{kg}\cdot 10.81\,\text{m/s}^2$$= 1297.2\,\text{N}$Answer
$1300\,\text{N}$
Or $1.3\,\text{kN}$, two significant figures like the velocity $2.0\,\text{m/s}$ and the time $2.0\,\text{s}$.
Reasonableness
The weight of the load is $120\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 1.18\,\text{kN}$. The force of the rope is $ma = 120\,\text{N}$ greater than this, so that the net force points up and stops the downward-moving load. If the rope held up only the weight, the load would keep going down at constant velocity.
- Exercise 5-12
Two children pull a cart of $30\,\text{kg}$ in a horizontal yard with horizontal ropes in opposite directions. One pulls with a force of $60\,\text{N}$ and the other with a force of $45\,\text{N}$. Friction is ignored.
What is the acceleration of the cart?
Answer
$0.50\,\text{m/s}^2$
Show the solution
Given
$m = 30\,\text{kg}$, forces $F_\text{a} = 60\,\text{N}$ and $F_\text{b} = 45\,\text{N}$ horizontally in opposite directions, no friction
Find
the acceleration $a$ of the cart in $\text{m/s}^2$
Principle
The acceleration is determined by the net force. Forces in opposite directions are subtracted from each other, and in the vertical direction the weight and the normal force from the ground cancel.
Free-body diagram
The cart as a rectangle, the positive direction the direction of the greater pull. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ points up from the bottom surface, equally long. The force of the rope $\overline{F}_\text{a}$ points forward from one side and $\overline{F}_\text{b}$ points backward from the other side, shorter.
Formula
$F_\text{a} - F_\text{b} = ma$, so
$a$$= \dfrac{F_\text{a} - F_\text{b}}{m}$Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$a$$= \dfrac{60\,\text{N} - 45\,\text{N}}{30\,\text{kg}}$$= \dfrac{15\,\text{N}}{30\,\text{kg}}$$= 0.50\,\text{m/s}^2$Answer
$0.50\,\text{m/s}^2$
In the direction of the greater pull, two significant figures like the data.
Reasonableness
The cart accelerates in the direction of the stronger puller, as in a tug of war. If both pulled with $60\,\text{N}$, the net force and the acceleration would be zero. The net force $15\,\text{N}$ is much smaller than either pull, so the acceleration is small.
- Exercise 5-13
On an air track a cart was pulled by a constant horizontal force, and the magnitude of the force was varied. The mass of the cart stayed the same, and the friction was so small that it can be ignored. For each force $F$ the acceleration $a$ of the cart was measured. The table is an example data set.
Draw the $(F, a)$ graph of the measurement points and fit a straight line through the origin to the points. Determine the slope of the line and use it to find the mass of the cart. Enter the mass of the cart in the answer box.
Acceleration of the cart with different pulling forces (example data) $F\,(\text{N})$ $a\,(\text{m/s}^2)$ 0.20 0.41 0.40 0.78 0.60 1.22 0.80 1.58 1.00 2.03 Answer
$0.50\,\text{kg}$
Show the solution
Given
the points $(F, a)$ of the table, for example $F = 1.00\,\text{N}$ and $a = 2.03\,\text{m/s}^2$; the mass is constant, friction is ignored
Find
the mass $m$ of the cart in $\text{kg}$
Principle
Friction is small, so the pulling force is the net force, and by Newton's second law $a = \dfrac{1}{m}F$. The acceleration is proportional to the force, so the line of the graph passes through the origin (no force, no acceleration) and its slope is $\dfrac{1}{m}$.
Reading the graph
The points are drawn in the $(F, a)$ coordinate system and a straight line through the origin is fitted between the points. From the line we read the point $F = 1.00\,\text{N}$, $a = 2.01\,\text{m/s}^2$. The slope is calculated using the origin and this point, not from two neighbouring measurement points.
Formula
$k = \dfrac{a}{F} = \dfrac{1}{m}$, so $m = \dfrac{1}{k}$
Conversions
The quantities are in SI units, and
$\dfrac{\text{m/s}^2}{\text{N}}$$= \dfrac{1}{\text{kg}}$so no conversions are needed.
Substitution
$k$$= \dfrac{2.01\,\text{m/s}^2 - 0\,\text{m/s}^2}{1.00\,\text{N} - 0\,\text{N}}$$= 2.01\,\tfrac{1}{\text{kg}}$$m$$= \dfrac{1}{2.01\,\tfrac{1}{\text{kg}}}$$= 0.4975\,\text{kg}$Answer
$0.50\,\text{kg}$
Two significant figures, because the slope is read from the graph to an accuracy of about two figures.
Reasonableness
At the measurement points $a : F$ is $0.41 : 0.20 \approx 2.05\,\tfrac{1}{\text{kg}}$ and $2.03 : 1.00 = 2.03\,\tfrac{1}{\text{kg}}$, so all the ratios are close to $2.0\,\tfrac{1}{\text{kg}}$. A mass of $0.5\,\text{kg}$ fits a small air-track cart. The points are not exactly on the line because of the random variation of the measurement.
- Exercise 5-14
A cart was pulled on a horizontal track every time with the same horizontal force. The mass $m$ of the cart was changed by adding weights to it, and with each mass the acceleration $a$ was measured. Friction was small and is ignored. The table is an example data set.
a) For each mass calculate the reciprocal $1/m$ and draw the $(1/m, a)$ graph. Fit a straight line through the origin to the points and determine its slope. What does the slope tell you?
b) Which quantities had to be kept constant for the measurement to be a fair test? Enter the magnitude of the pulling force in the answer box.
Acceleration of the cart with different masses (example data) $m\,(\text{kg})$ $a\,(\text{m/s}^2)$ 0.20 6.1 0.30 3.9 0.40 3.1 0.50 2.4 0.60 2.0 Answer
$1.2\,\text{N}$
Show the solution
Given
the pulling force is constant, the points $(m, a)$ of the table, for example $m = 0.20\,\text{kg}$ and $a = 6.1\,\text{m/s}^2$; friction is ignored
Find
the magnitude of the pulling force $F$ in $\text{N}$
Principle
With a constant force $a = F \cdot \dfrac{1}{m}$, so the acceleration is proportional to the reciprocal of the mass $\dfrac{1}{m}$. With the reciprocal, the curve $(m, a)$ becomes a straight line, and the slope of the line is the force $F$. The line passes through the origin, because with a very large mass the acceleration would be almost zero. The condition for the slope and the straight line to make sense is that only the mass changes: the pulling force, the surface and the friction must stay the same.
Reading the graph
The reciprocals are $5.0$, $3.3$, $2.5$, $2.0$ and $1.7\,\tfrac{1}{\text{kg}}$. The points $(1/m, a)$ lie on a straight line, and from the line we read the points $(0, 0)$ and $(5.0\,\tfrac{1}{\text{kg}}, 6.0\,\text{m/s}^2)$.
Formula
$a = F \cdot \dfrac{1}{m}$, so the slope $k = \dfrac{\Delta a}{\Delta(1/m)} = F$
Conversions
The quantities are in SI units, and
$\text{kg}\cdot\dfrac{\text{m}}{\text{s}^2}$$= \text{N}$so no conversions are needed.
Substitution
$F$$= k$$= \dfrac{6.0\,\text{m/s}^2 - 0\,\text{m/s}^2}{5.0\,\tfrac{1}{\text{kg}} - 0\,\tfrac{1}{\text{kg}}}$$= 1.2\,\text{N}$Answer
a) the slope is the pulling force $F$; b) the pulling force, the track and the friction were kept constant, and only the mass changed. The magnitude of the force is $1.2\,\text{N}$, two significant figures like the readings of the accelerations.
Reasonableness
At every point $ma$ is close to the same value: $0.20\,\text{kg} \cdot 6.1\,\text{m/s}^2 = 1.22\,\text{N}$ and $0.50\,\text{kg} \cdot 2.4\,\text{m/s}^2 = 1.20\,\text{N}$. If the mass doubles, the acceleration halves: $0.30\,\text{kg}$ gives $3.9\,\text{m/s}^2$ and $0.60\,\text{kg}$ gives $2.0\,\text{m/s}^2$.
- Exercise 5-15
A crane lowers a concrete element of mass $820\,\text{kg}$ onto a quay. The element moves downward and slows down uniformly, and the magnitude of its acceleration is $1.20\,\text{m/s}^2$. Air resistance is ignored.
A student calculated the tension in the rope like this: “The element moves downward, so the acceleration points down. Thus $mg - T = ma$, from which $T = m(g - a) = 820\,\text{kg}\cdot(9.81\,\text{m/s}^2 - 1.20\,\text{m/s}^2) = 7.06\,\text{kN}$.”
What is the error in the student's solution? Calculate the correct tension in the rope. Enter the magnitude of the tension in the rope in the answer box.
Answer
$9030\,\text{N}$
Show the solution
Given
$m = 820\,\text{kg}$, magnitude of the acceleration $a = 1.20\,\text{m/s}^2$, the element moves down and slows down; $g = 9.81\,\text{m/s}^2$
Find
the tension $T$ in the rope in $\text{N}$ and the error in the student's solution
Principle
By Newton's second law $\Sigma\overline{F} = m\overline{a}$ the net force points in the direction of the acceleration, not the direction of the velocity. The element moves down but slows down, so its speed decreases and the acceleration points up. The student chose the direction of motion, that is, down, as the direction of the acceleration. The rope must pull upward harder than the weight so that the net force points up.
Free-body diagram
The concrete element as a rectangle, the positive direction up. The tension $\overline{T}$ of the rope points up from the top surface and the weight $\overline{G}$ points down from the centre. The arrow $\overline{T}$ is longer than $\overline{G}$. The velocity points down and the acceleration up.
Formula
$T - mg = ma$, so $T = m(g + a)$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$T$$= 820\,\text{kg}\cdot(9.81\,\text{m/s}^2 + 1.20\,\text{m/s}^2)$$= 820\,\text{kg}\cdot 11.01\,\text{m/s}^2$$= 9028.2\,\text{N}$Answer
$9030\,\text{N}$
Or $9.03\,\text{kN}$, three significant figures like the acceleration $1.20\,\text{m/s}^2$. The student's error was to take the direction of motion, down, as the direction of the acceleration: the acceleration points up, because the element slows down.
Reasonableness
The weight is $820\,\text{kg}\cdot 9.81\,\text{m/s}^2 = 8044.2\,\text{N}$. For the slowing acceleration an additional $820\,\text{kg}\cdot 1.20\,\text{m/s}^2 = 984\,\text{N}$ is needed, and the sum is $9028.2\,\text{N}$. The student's result $7.06\,\text{kN}$ would be smaller than the weight, and then the element would accelerate down instead of slowing down.
Level 3 · challenging 5 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 5-16
A student of mass $60.0\,\text{kg}$ stands on a bathroom scale in an elevator. The scale measures the force with which it supports the student and shows it in kilograms, that is, divides the force by $9.81\,\text{m/s}^2$. When the elevator starts to move, the scale shows $54.0\,\text{kg}$ for a moment.
What is the acceleration of the elevator, and in which direction does it point?
Answer
$0.98\,\text{m/s}^2$
Show the solution
Given
mass of the student $m = 60.0\,\text{kg}$, reading of the scale $54.0\,\text{kg}$, so the normal force of the scale is $N = 54.0\,\text{kg}\cdot g$; $g = 9.81\,\text{m/s}^2$
Find
the acceleration $a$ of the elevator (and the student) in $\text{m/s}^2$ and its direction
Principle
The scale does not measure mass but the normal force with which it pushes on the student. The student accelerates with the elevator, so by Newton's second law the difference between the normal force and the weight is $ma$.
Free-body diagram
The student as a rectangle, the positive direction up. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ from the scale points up from the bottom surface. The reading is smaller than the mass, so $\overline{N}$ is shorter than $\overline{G}$ and the net force points down.
Formula
$N - mg = ma$, so $a = \dfrac{N - mg}{m}$
Conversions
The reading of the scale is converted to a force:
$N$$= 54.0\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 529.74\,\text{N}$Substitution
$a$$= \dfrac{529.74\,\text{N} - 60.0\,\text{kg}\cdot 9.81\,\text{m/s}^2}{60.0\,\text{kg}}$$= \dfrac{529.74\,\text{N} - 588.6\,\text{N}}{60.0\,\text{kg}}$$= -0.981\,\text{m/s}^2$Answer
$0.98\,\text{m/s}^2$
Downward, two significant figures, because the difference of the readings, $6.0\,\text{kg}$, is known to two figures.
Reasonableness
The scale shows ten percent too little, so the acceleration is a tenth of the free-fall acceleration, $0.1\cdot 9.81\,\text{m/s}^2$. The minus sign means an acceleration downward: the elevator starts moving down. The same reading would also be shown if an elevator moving upward were slowing down.
- Exercise 5-17
An empty timber truck of mass $10\,\text{t}$ accelerates uniformly from rest to a velocity of $36\,\text{km/h}$ in $8.0\,\text{s}$. Fully loaded, its mass is $25\,\text{t}$. Assume that the net force during the acceleration is the same as for the empty truck.
How long does it take the loaded truck to accelerate from rest to $36\,\text{km/h}$?
Answer
$20\,\text{s}$
Show the solution
Given
mass of the empty truck $m_1 = 10\,\text{t}$, of the loaded truck $m_2 = 25\,\text{t}$, change in velocity $\Delta v = 36\,\text{km/h}$, time of the empty truck $\Delta t_1 = 8.0\,\text{s}$, the same net force
Find
the acceleration time $\Delta t_2$ of the loaded truck in $\text{s}$
Principle
The net force is obtained from the acceleration of the empty truck by Newton's second law. The same net force gives a greater mass a smaller acceleration, and with a smaller acceleration the same change in velocity takes longer.
Formula
$\Sigma F$$= m_1 \dfrac{\Delta v}{\Delta t_1}$$\Sigma F$$= m_2 \dfrac{\Delta v}{\Delta t_2}$so
$\Delta t_2$$= \dfrac{m_2 \Delta v}{\Sigma F}$Conversions
$36\,\text{km/h}$$= 36 : 3.6\,\text{m/s}$$= 10\,\text{m/s}$$10\,\text{t} = 10\,000\,\text{kg}$ and $25\,\text{t} = 25\,000\,\text{kg}$
Substitution
$\Sigma F$$= 10\,000\,\text{kg}\cdot\dfrac{10\,\text{m/s}}{8.0\,\text{s}}$$= 12\,500\,\text{N}$$\Delta t_2$$= \dfrac{25\,000\,\text{kg}\cdot 10\,\text{m/s}}{12\,500\,\text{N}}$$= 20\,\text{s}$Answer
$20\,\text{s}$
Two significant figures like the time $8.0\,\text{s}$.
Reasonableness
The mass grows $2.5$ times, so the acceleration becomes $2.5$ times smaller and the time becomes $2.5$ times longer: $2.5\cdot 8.0\,\text{s} = 20\,\text{s}$. This explains why a loaded truck is left behind at the traffic lights.
- Exercise 5-18
A child slides on a sled on horizontal snow. The total mass of the sled and the child is $30\,\text{kg}$, and the velocity is $4.0\,\text{m/s}$ at the start. Nobody pulls or pushes the sled. The friction force from the snow is $18\,\text{N}$, and air resistance is not taken into account.
How long does it take before the sled stops?
Answer
$6.7\,\text{s}$
Show the solution
Given
$m = 30\,\text{kg}$, initial velocity $v_0 = 4.0\,\text{m/s}$, final velocity $v = 0$, friction force $F_\mu = 18\,\text{N}$
Find
the time $\Delta t$ in $\text{s}$ after which the sled stops
Principle
The sled is neither pulled nor pushed, so the only horizontal force is the friction, which points against the motion. The friction is constant, so by Newton's second law the deceleration is constant, and the stopping time follows from the change in velocity.
Free-body diagram
The sled and the child as a rectangle, the positive direction the direction of motion. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ from the snow points up from the bottom surface, equally long. The friction force $\overline{F}_\mu$ from the snow points backward from the back corner of the bottom surface. There is no force pushing forward.
Formula
$-F_\mu = ma$ and $a = \dfrac{v - v_0}{\Delta t}$, so $\Delta t = \dfrac{m(v_0 - v)}{F_\mu}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$\Delta t$$= \dfrac{30\,\text{kg}\cdot(4.0\,\text{m/s} - 0)}{18\,\text{N}}$$= \dfrac{120\,\text{kg}\cdot\text{m/s}}{18\,\text{N}}$$= 6.67\,\text{s}$Answer
$6.7\,\text{s}$
Two significant figures like the data $4.0\,\text{m/s}$, $18\,\text{N}$ and $30\,\text{kg}$.
Reasonableness
The deceleration is $18\,\text{N} : 30\,\text{kg} = 0.60\,\text{m/s}^2$, so the speed decreases by $0.6\,\text{m/s}$ every second, and $4.0\,\text{m/s}$ takes about seven seconds. The sled keeps moving forward the whole time even though the net force points backward: it just slows down.
- Exercise 5-19
Students investigate how the acceleration $a$ of a cart pulled along a table depends on the horizontal pulling force $F$. The mass of the cart stayed the same. A constant friction force from the surface acts on the moving cart. The table is an example data set.
a) Draw the $(F, a)$ graph of the measurement points and fit a straight line to the points. Justify why the line does not pass through the origin.
b) Determine the mass of the cart from the slope of the line.
c) Determine the magnitude of the friction force from the point where the line crosses the $F$-axis. Enter the magnitude of the friction force in the answer box.
Acceleration of the cart with different pulling forces (example data) $F\,(\text{N})$ $a\,(\text{m/s}^2)$ 2.0 0.40 3.0 0.91 4.0 1.42 5.0 1.88 6.0 2.41 Answer
$1.2\,\text{N}$
Show the solution
Given
the points $(F, a)$ of the table for $F = 2.0 \ldots 6.0\,\text{N}$; the mass is constant, the friction force is constant
Find
the magnitude of the friction force $F_\mu$ in $\text{N}$ (and also the mass $m$ of the cart)
Principle
On the moving cart the pulling force acts in the direction of motion and friction against it, so by Newton's second law $F - F_\mu = ma$, that is, $a = \dfrac{1}{m}F - \dfrac{F_\mu}{m}$. The graph is a straight line whose slope is $\dfrac{1}{m}$. The line does not pass through the origin, because part of the pulling force is used to overcome friction: the acceleration is zero when $F = F_\mu$, so the line crosses the $F$-axis at the value of the friction force.
Reading the graph
A straight line is fitted between the points. From the line we read two points far from each other: $(2.0\,\text{N}, 0.41\,\text{m/s}^2)$ and $(6.0\,\text{N}, 2.40\,\text{m/s}^2)$. The slope is calculated from them, and the crossing point on the $F$-axis is obtained from the first point with the slope.
Formula
$k$$= \dfrac{\Delta a}{\Delta F}$$= \dfrac{1}{m}$so $m = \dfrac{1}{k}$, and at the crossing point $a = 0$, so $F_\mu = F_1 - \dfrac{a_1}{k}$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$k$$= \dfrac{2.40\,\text{m/s}^2 - 0.41\,\text{m/s}^2}{6.0\,\text{N} - 2.0\,\text{N}}$$= 0.4975\,\tfrac{1}{\text{kg}}$$m$$= \dfrac{1}{0.4975\,\tfrac{1}{\text{kg}}}$$= 2.01\,\text{kg}$$F_\mu$$= 2.0\,\text{N} - \dfrac{0.41\,\text{m/s}^2}{0.4975\,\tfrac{1}{\text{kg}}}$$= 1.18\,\text{N}$Answer
$1.2\,\text{N}$
($1.18\,\text{N}$ before rounding), two significant figures. The mass of the cart is about $2.0\,\text{kg}$. The points are not exactly on the line because of the random variation of the measurement, so the result is an estimate.
Reasonableness
When $F = 4.0\,\text{N}$, the net force is
$4.0\,\text{N} - 1.18\,\text{N}$$= 2.82\,\text{N}$$a$$= 2.82\,\text{N} : 2.01\,\text{kg}$$= 1.40\,\text{m/s}^2$which matches the value $1.42\,\text{m/s}^2$ of the table. Friction is small compared with the weight $2.01\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 19.7\,\text{N}$, about $6\,\%$ of it, which fits a smooth surface.
- Exercise 5-20
A person of mass $70\,\text{kg}$ stands in an elevator. The elevator starts upward from rest and moves in three phases. The accelerations of phases 1 and 2 have been measured with an acceleration sensor (example data), and the positive direction is up.
In phase 3 the elevator slows down uniformly and stops at the moment $t = 13.0\,\text{s}$.
a) Calculate the greatest velocity of the elevator.
b) Calculate the acceleration of the elevator in phase 3.
c) Draw the free-body diagram of the person in phase 3 and calculate the normal force from the floor on the person. Compare it with the weight of the person and justify why the normal force differs from the weight. Enter the normal force of part c) in the answer box.
Acceleration of the elevator in phases 1 and 2 (example data) Phase $t_{\text{start}}\,(\text{s})$ $t_{\text{end}}\,(\text{s})$ $a\,(\text{m/s}^2)$ 1 0.0 3.0 1.0 2 3.0 9.0 0.0 Answer
$630\,\text{N}$
Show the solution
Given
$m = 70\,\text{kg}$, phase 1: $0 \ldots 3.0\,\text{s}$, $a_1 = 1.0\,\text{m/s}^2$; phase 2: $3.0 \ldots 9.0\,\text{s}$, $a_2 = 0$; phase 3: $9.0 \ldots 13.0\,\text{s}$, uniform slowing down to a stop; initial velocity $0$; $g = 9.81\,\text{m/s}^2$
Find
the greatest velocity $v_{\max}$, the acceleration $a_3$ and the normal force $N$ from the floor in phase 3
Principle
In uniformly accelerated motion the change of velocity is $\Delta v = a\Delta t$. The elevator accelerates only in phase 1, so the greatest velocity is at the end of phase 1 and it is kept in phase 2. In phase 3 the elevator slows down to a stop. The forces on the person are from the Earth (weight) and from the floor (normal force), and by Newton's second law their sum is $ma$.
Free-body diagram
The person as a rectangle, the positive direction up. The weight $\overline{G}$ points down from the centre and the normal force $\overline{N}$ from the floor points up from the bottom surface. The arrow $\overline{N}$ is shorter than $\overline{G}$, because the acceleration points down. The velocity points up.
Formula
$v_{\max}$$= a_1\Delta t_1$$a_3$$= \dfrac{0 - v_{\max}}{\Delta t_3}$$N - mg = ma_3$, so $N = m(g + a_3)$
Conversions
The quantities are in SI units, so no conversions are needed. The duration of phase 3 is
$\Delta t_3$$= 13.0\,\text{s} - 9.0\,\text{s}$$= 4.0\,\text{s}$Substitution
$v_{\max}$$= 1.0\,\text{m/s}^2 \cdot 3.0\,\text{s}$$= 3.0\,\text{m/s}$$a_3$$= \dfrac{0 - 3.0\,\text{m/s}}{4.0\,\text{s}}$$= -0.75\,\text{m/s}^2$The normal force is
$N$$= 70\,\text{kg}\cdot(9.81\,\text{m/s}^2 - 0.75\,\text{m/s}^2)$$= 70\,\text{kg}\cdot 9.06\,\text{m/s}^2$$= 634.2\,\text{N}$Answer
$630\,\text{N}$
($634\,\text{N}$ before rounding), two significant figures like the mass $70\,\text{kg}$ and the acceleration. The greatest velocity is $3.0\,\text{m/s}$ and the acceleration in phase 3 is $0.75\,\text{m/s}^2$ downward. The normal force is smaller than the weight $mg = 686.7\,\text{N}$, because the net force and the acceleration point down.
Reasonableness
The elevator stops: $3.0\,\text{m/s} - 0.75\,\text{m/s}^2 \cdot 4.0\,\text{s} = 0$. The difference between the weight and the normal force is
$686.7\,\text{N} - 634.2\,\text{N}$$= 52.5\,\text{N}$$= 70\,\text{kg}\cdot 0.75\,\text{m/s}^2$so the forces agree with the equation of motion. In phase 1 the normal force would correspondingly be greater than the weight, $70\,\text{kg}\cdot 10.81\,\text{m/s}^2 = 756.7\,\text{N}$, and in the steady phase 2 equal to the weight.