Applied lesson: lift, parachute and sledding hill
Why can two sleds stop equally far from the top of a hill, although one goes down a steep slope and the other a gentle one?
Prior knowledge
- The net force points in the direction of the acceleration, and $\Sigma\overline{F} = m\overline{a}$chapter Newton's second law
- On a slope the normal force is $mg\cos\alpha$ and the kinetic friction is $\mu N$chapter Weight, friction and the inclined plane
- The change in mechanical energy is the work done by forces other than the weightchapter Conservation of mechanical energy and the work–energy principle
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 2.
Think about it
A physics group takes a bathroom scale into the lift of a shopping centre. When the lift starts up, the reading rises for a moment, and when the lift stops, it falls. Through the window you can see a skydiver who descends under a parachute at a steady speed, although the weight pulls the skydiver down all the time. On a sledding hill two children go down different sides of the same hill: one a steep, short slope, the other a gentle, long slope. The one who came down the steep side slides much farther on the flat, but when the stopping points are measured, both are equally far from the top of the hill.
Familiar forces act in all three situations. When should you solve a situation with forces and when with energy, and why do the sleds stop equally far from the top?
Optional Review: the work–energy principle
Predict
Make your predictions before investigating 2 questions
Prediction 1: a scale in a lift
Prediction 2: which is faster?
Investigate
A force sensor in a lift
Made-up example data
A group takes a force-sensor plate that measures the normal force in newtons into a lift. A student of mass $60.0\,\text{kg}$ stands on the plate. The lift starts up from the ground floor, travels for a while at a steady speed and stops at an upper floor. The graph shows the reading of the plate as a function of time. The numbers are invented example data.
First draw the free-body diagram of the student in your notebook. Then read the normal force at the start from the graph and calculate the acceleration of the lift. Give the answer in $\text{m/s}^2$.
Hint 1: a guiding question
Which two forces act on the student, and which one is larger at the start?
Hint 2: what is the connection
Choose up as the positive direction: $N - mg = ma$. Read $N$ from the graph between $1.0\,\text{s}$ and $3.0\,\text{s}$.
Hint 3: an intermediate step
Divide the difference by the mass.
Show the solution
Given
mass of the student $m = 60.0\,\text{kg}$; from the graph the normal force in the start phase $N = 650\,\text{N}$ (from $1.0\,\text{s}$ to $3.0\,\text{s}$); $g = 9.81\,\text{m/s}^2$
Find
the acceleration $a$ of the lift in the start phase in $\text{m/s}^2$
Principle
The student stands still in the lift, so the student's acceleration is the acceleration of the lift. By Newton's second law it is determined by the net force acting on the student, the sum of the weight and the normal force.
Free-body diagram
The student as a rectangle, positive direction up. The weight $\overline{G}$ from the centre down and the normal force $\overline{N}$ of the plate from the bottom surface up, longer than the weight. The acceleration up is marked separately and not as a force.
Reading the graph
In the start phase, from $1.0\,\text{s}$ to $3.0\,\text{s}$, the curve is horizontal and meets the grid line at $650\,\text{N}$. At rest the reading is about $589\,\text{N}$, so at the start the normal force is larger than the weight.
Formula
$N - mg = ma$, so $a = \dfrac{N}{m} - g$
Conversions
The mass is in kilograms and the force in newtons, and $\tfrac{\text{N}}{\text{kg}} = \tfrac{\text{m}}{\text{s}^2}$, so no conversions are needed.
Substitution
Answer
$1.02\,\text{m/s}^2$
Upwards, three significant figures like the mass $60.0\,\text{kg}$ and the $650\,\text{N}$ read from the graph.
Reasonableness
The normal force is about $60\,\text{N}$ larger than the weight, and $60\,\text{N}$ accelerates a $60\,\text{kg}$ student by about $1\,\text{m/s}^2$. The acceleration of a lift is typically about $1\,\text{m/s}^2$, and it is only slightly felt in the legs.
Three hills of equal height
Values calculated by a model
The table shows, calculated with a model, how a sled slides from three hills of the same height, $h = 6.0\,\text{m}$. The sled starts from rest at the top, the coefficient of kinetic friction is $\mu = 0.10$ both on the slope and on the flat, air resistance is ignored and no more energy is lost in the bend at the foot of the slope. The angle $\alpha$ is the angle of inclination of the slope, $x_1$ the horizontal length of the slope, $v$ the speed of the sled at the foot of the slope and $x_2$ the distance the sled slides on the flat before it stops. The numbers are values calculated with a model, not measurements.
| $\alpha$ | $x_1$ (m) | $v$ (m/s) | $x_2$ (m) |
|---|---|---|---|
| $30^\circ$ | $10.4$ | $9.87$ | $49.6$ |
| $20^\circ$ | $16.5$ | $9.24$ | $43.5$ |
| $10^\circ$ | $34.0$ | $7.14$ | $26.0$ |
Conclusion
Use the graph and the table to explain two things.
What does the reading of a scale in a lift depend on? Why is the sled from the steep slope faster at the foot but still stops almost as far from the top as the sled from the gentle slope?
Model answer
A scale measures the normal force with which it pushes the person standing on it. In the graph the reading is $650\,\text{N}$ at the start, about $589\,\text{N}$ at a steady speed, as at rest, and $550\,\text{N}$ when stopping. The reading is larger than the weight when the acceleration of the lift points up and smaller when it points down. The speed does not affect the reading. In the table the sled from the steep slope is fastest at the foot, $9.87\,\text{m/s}$, because on a steep slope the normal force and the friction are smaller and the slope is shorter, so friction does less negative work. It slides farthest on the flat, but the horizontal distance on its slope is the shortest. In the values of the model $x_1 + x_2$ is $60.0\,\text{m}$ in every row, so the inclination does not affect how far from the top the sled stops.
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
First the free-body diagram, then the tool
The mechanics problems of this module are solved with two tools, and both start from the same place. First you choose the body and draw its free-body diagram, as in the chapter Free-body diagrams and the combined effect of forces. The diagram shows which forces act and in which directions. Then you choose the tool according to what is asked. When the question is about a force or an acceleration at some moment, you write the equation of motion (chapter Newton's second law). When the question is about the speed at some point or the change in energy over some distance, you use the work–energy principle (chapter Conservation of mechanical energy and the work–energy principle). Even then you need the free-body diagram, because it shows which forces do work.
Lift: a scale measures the normal force
A bathroom scale does not measure the weight but the normal force $N$ with which it pushes the person standing on it upwards. The reading is shown in kilograms, because the scale divides the force by $9.81\,\text{m/s}^2$. Only the weight and the normal force act on a person standing still in a lift. When the positive direction is up, the equation of motion is $N - mg = ma$, so $N = m(g + a)$. When the lift starts up and accelerates at $1.0\,\text{m/s}^2$, the normal force on a $60.0\,\text{kg}$ person is $649\,\text{N}$, rounded $650\,\text{N}$ as in the graph, and the scale shows $66\,\text{kg}$. At a steady speed $a = 0$, and the scale shows the same as at home, whether the lift moves up or down. So the reading depends on the acceleration, not on the speed.
Parachute: at terminal velocity the mechanical energy decreases
A skydiver descending under a parachute soon falls at terminal velocity, that is, at a steady speed (chapter Air resistance and falling through air). By the equation of motion the acceleration is zero, so the air resistance is equal to the weight, $F_\text{d} = mg$. In terms of energy, the kinetic energy does not change, but the potential energy decreases all the time. By the work–energy principle the change in mechanical energy is the work done by air resistance. It is negative, because air resistance points against the motion: over a height $h$, $W_\text{d} = -F_\text{d} h = -mgh$. When an $80\,\text{kg}$ skydiver descends $100\,\text{m}$, about $78\,\text{kJ}$ of mechanical energy is changed into motion of the air and finally into internal energy. Each tool tells the same thing in its own way.
Sledding hill: friction takes energy all the way
A sled slides down a straight slope. The free-body diagram shows that the component of the weight perpendicular to the slope and the normal force cancel each other, so $N = mg\cos\alpha$, as in the chapter Weight, friction and the inclined plane. The kinetic friction is then $F_\mu = \mu mg\cos\alpha$. The normal force does no work, because it is perpendicular to the motion. On the slope the weight does work $mgh$, and friction does negative work $-F_\mu L$, when the length of the slope is $L$. On a hill of the same height, the normal force and the friction are smaller on a steep slope and the slope is shorter, so friction takes less energy and the sled is faster at the foot. On the flat the sled slides until friction has turned all the kinetic energy into internal energy.
Equation of motion and work–energy principle
The first tool of an applied problem is Newton's second law:
$$\Sigma\overline{F} = m\overline{a}$$
where $\Sigma\overline{F}$ is the sum of the forces acting on the body, that is, the net force ($\text{N}$), $m$ is the mass of the body ($\text{kg}$) and $\overline{a}$ its acceleration ($\text{m/s}^2$). The law is written from the free-body diagram in components, $\Sigma F_x = ma_x$ and $\Sigma F_y = ma_y$, once the axes and the positive directions have been chosen. The second tool is 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, $E_\text{k} = \tfrac12 mv^2$ is the kinetic energy, $E_\text{p} = mgh$ the potential energy from the chosen zero level and $W_\text{other}$ the work done by forces other than the weight ($\text{J}$). The work–energy principle needs neither the acceleration nor the time, so it fits when the question is about the speed at a given point.
The work done by friction depends on the horizontal distance
On a straight slope with an angle of inclination $\alpha$ and a length $L$, the kinetic friction is $\mu mg\cos\alpha$ and the work it does is $-\mu mg\cos\alpha \cdot L$. The figure shows that $L\cos\alpha$ is the horizontal length $x_1$ of the slope. On the flat $N = mg$, and the work done by friction over a distance $x_2$ is $-\mu mg x_2$. In both cases the work done by kinetic friction is
$$W_\mu = -\mu m g x$$
where $\mu$ is the coefficient of kinetic friction, $m$ the mass ($\text{kg}$), $g$ the acceleration of free fall ($\text{m/s}^2$) and $x$ the horizontal length of the distance travelled ($\text{m}$). The result requires that the surface is straight, that the coefficient of friction is the same all the way and that only the weight and the normal force act perpendicular to the surface, so the sled is not, for example, pulled with a rope.
Why the sleds stop equally far away
A sled starts from rest at a height $h$, goes down the slope and stops on the flat. When the zero level is chosen at the flat, the work–energy principle gives $mgh - \mu mg(x_1 + x_2) = 0$. The mass cancels, and the whole horizontal distance of the sled from the top to the stopping point is
$$x_1 + x_2 = \frac{h}{\mu}$$
The angle of inclination does not appear in the result. A sled coming down the steep side travels a short horizontal distance on the slope, so it has more distance left on the flat. A sled coming down the gentle side uses a larger part of the horizontal distance on the slope. This answers the question at the start. The model assumes that the coefficient of friction is the same all the way, that air resistance is small and that the bend at the foot of the slope is short. In a sharp bend the normal force and the friction grow for a moment, so a real sled stops a little earlier below a steep hill. In addition, the slope must be steep enough for the sled to slide: $\tan\alpha > \mu$.
Steps of the solution
- Choose the body and draw its free-body diagram: for every force its cause and direction.
- Decide the tool: the force or acceleration at a moment from the equation of motion, the speed at a given point from the work–energy principle.
- For the equation of motion choose the axes and the positive directions. For the work–energy principle choose the zero level and write the work of the other forces with their signs: a force opposing the motion does negative work, and a force perpendicular to the motion does no work.
- Solve the asked quantity in symbols, substitute only then, and check the result with a limiting case, for example without friction.
Common misconception
Many think that a scale in a lift shows more because the lift moves upwards. The reading depends on the acceleration, not on the speed: in a lift moving steadily up or down the scale shows the same as at home, and in a lift moving up that slows down to stop it shows less.
Another mistake is to think that at terminal velocity the mechanical energy is conserved, because the speed does not change. The kinetic energy stays the same, but the potential energy decreases, and the negative work done by air resistance changes it into other forms of energy. A third mistake is to write the friction force on a slope as $\mu mg$. On a slope the normal force is $mg\cos\alpha$, so the friction force is $\mu mg\cos\alpha$, and the work done by friction is $-\mu mg$ times the horizontal distance, not times the length of the slope.
Summary
- A mechanics problem always starts with choosing the body and drawing its free-body diagram. The force or acceleration at a moment comes from the equation of motion $\Sigma\overline{F} = m\overline{a}$, the speed at a given point from the work–energy principle.
- In a lift a scale measures the normal force $N = m(g + a)$, when the positive direction is up. The reading depends on the acceleration, not on the speed.
- At terminal velocity $F_\text{d} = mg$, and the work done by air resistance is $-mgh$: the mechanical energy decreases although the speed does not change.
- On a straight slope $N = mg\cos\alpha$, and the work done by kinetic friction is $W_\mu = -\mu mgx$, where $x$ is the horizontal distance. A sled that starts from rest at a height $h$ stops at a horizontal distance $h/\mu$ from the top, when the coefficient of friction is the same all the way.
- When a body slides along a stationary surface, the normal force does no work, because it is perpendicular to the motion. In a lift the normal force of the floor is in the direction of the motion and does work.
Optional Return to your predictions
Compare your predictions with the graph and the table. What surprised you most? Write in your own words why the work done by friction depends on the horizontal distance and not on the length of the slope.
Would you change your prediction?
Apply
Worked example
Problem
A child and a sled, together $32.0\,\text{kg}$, start from rest at the top of a straight slope. The length of the slope is $25.0\,\text{m}$, the angle of inclination $20.0^\circ$ and the coefficient of kinetic friction between the sled and the snow $0.200$. How large is the speed of the sled at the foot of the slope? Air resistance is ignored.
Step 1/8
Given and asked. $m = 32.0\,\text{kg}$, $L = 25.0\,\text{m}$, $\alpha = 20.0^\circ$, $\mu = 0.200$, initial speed $v_1 = 0$ and $g = 9.81\,\text{m/s}^2$. We want the speed $v_2$ at the foot of the slope.
Step 2/8
Principle. The question is about the speed at a given point, so we use the work–energy principle. We choose the sled and the child together as the body. The weight does work, which the work–energy principle takes into account as a change in potential energy. The normal force does no work, because it is perpendicular to the motion. Kinetic friction points against the motion and does negative work $W_\text{other} = -F_\mu L$. The friction force comes from the free-body diagram.
Step 3/8
Free-body diagram. The sled is drawn as a rectangle along the slope. The forces on it are from the Earth (the weight $\overline{G}$), from the snow (the normal force $\overline{N}$) and the kinetic friction $\overline{F}_\mu$ of the snow, which points up the slope, because the sled slides down. The axes are chosen so that $x$ points down the slope and $y$ perpendicularly out of the slope. In the $y$ direction the sled does not accelerate, so $N - mg\cos\alpha = 0$ and $F_\mu = \mu N = \mu mg\cos\alpha$.
Step 4/8
Formula solved. The zero level is chosen at the foot of the slope, so at the start $E_\text{p,1} = mgh$, where $h = L\sin\alpha$, and at the end $E_\text{p,2} = 0$. The speed is solved from the work–energy principle:
$$\begin{aligned} &mgL\sin\alpha - \mu mg\cos\alpha \cdot L \\ \quad &= \tfrac12 mv_2^2 \\ v_2 &= \sqrt{2gL(\sin\alpha - \mu\cos\alpha)} \end{aligned}$$
Step 5/8
Unit conversions. Not needed: the quantities are in SI units, the coefficient of friction has no unit and the angle is in degrees with the calculator in degree mode.
Step 6/8
Substitution. First we calculate the expression in brackets and then the speed:
$$\begin{aligned} &\sin 20.0^\circ - 0.200\cdot\cos 20.0^\circ \\ \quad &\approx 0.3420 - 0.1879 = 0.1541 \\ v_2 &= \sqrt{2\cdot 9.81\cdot 25.0\cdot 0.1541}\,\tfrac{\text{m}}{\text{s}} \\ \phantom{v_2} &\approx 8.694\,\text{m/s} \end{aligned}$$
Step 7/8
Result. The speed of the sled at the foot of the slope is $8.69\,\text{m/s}$, or about $31\,\text{km/h}$, to three significant figures like the length of the slope $25.0\,\text{m}$, the angle $20.0^\circ$ and the coefficient of friction $0.200$.
Step 8/8
Reasonableness. Without friction the speed would be
so friction reduces it, as it should. The same result comes from the equation of motion: in the $x$ direction $mg\sin\alpha - \mu mg\cos\alpha = ma$, so $a \approx 1.51\,\text{m/s}^2$, and from the result $aL = \tfrac12 v_2^2$ of the chapter Conservation of mechanical energy and the work–energy principle we get $v_2^2 \approx 75.6\,\text{m}^2/\text{s}^2$ as above. The mass cancelled, so a heavier and a lighter sled are equally fast at the foot, when the coefficient of friction is the same.
Draw it yourself: a suitcase in a lift
A hook hangs from the ceiling of a lift, and from it hangs a spring scale with an $18.0\,\text{kg}$ suitcase on its hook. The lift starts downwards and accelerates. Draw the free-body diagram of the suitcase on paper: the suitcase as a rectangle and every force at its point of application. Mark the directions of the velocity and the acceleration separately next to the diagram. Then write the equation of motion when the positive direction is down.
Does the scale show more or less than when the lift is at rest?
Model diagram
Only the Earth (the weight $\overline{G}$) and the scale, whose hook holds the suitcase (the tension $\overline{T}$), act on the suitcase. The lift and the ceiling act on the suitcase only through the scale, so they get no forces of their own. The lift moves down and accelerates, so the velocity and the acceleration point down. The net force must point down, so the tension is smaller than the weight. When the positive direction is down, the equation of motion is $mg - T = ma$, from which $T = m(g - a)$. The scale shows less than at rest. For example, when $a = 1.5\,\text{m/s}^2$, $T = 18.0\,\text{kg}\cdot 8.31\,\text{m/s}^2 \approx 150\,\text{N}$, when the weight is $177\,\text{N}$.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Apply to a new situation
A skydiver whose mass with equipment is $78.0\,\text{kg}$ descends under an open parachute at a steady speed of $5.0\,\text{m/s}$.
How much mechanical energy is changed into other forms of energy as the skydiver descends $125\,\text{m}$? First draw the free-body diagram. Give the answer in kilojoules.
Hint 1: a guiding question
What happens to the kinetic energy and the potential energy of the skydiver when the speed is constant?
Hint 2: what is the connection
The speed is constant, so $F_\text{d} = mg$. The work done by air resistance is $W_\text{d} = -F_\text{d} h$, and it is the change in mechanical energy.
Hint 3: an intermediate step
$mgh = 78.0\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot 125\,\text{m}$. Change the joules into kilojoules.
Show the solution
Given
$m = 78.0\,\text{kg}$, the speed is constant $v = 5.0\,\text{m/s}$, the change in height $h = 125\,\text{m}$; $g = 9.81\,\text{m/s}^2$
Find
the mechanical energy changed into other forms of energy $|\Delta E_\text{mech}|$ in $\text{kJ}$
Principle
By the work–energy principle the change in mechanical energy is the work done by forces other than the weight, here the work $W_\text{d}$ done by air resistance. The magnitude of the air resistance comes from the equation of motion: the speed is constant, so the net force is zero.
Free-body diagram
The skydiver with the parachute as a rectangle, positive direction down. The weight $\overline{G}$ from the centre down and the air resistance $\overline{F}_\text{d}$ from the top surface up, equally long. The velocity down is marked separately and not as a force.
Formula
$mg - F_\text{d} = 0$, so $F_\text{d} = mg$. The air resistance points against the motion, so $W_\text{d} = -F_\text{d} h = -mgh$, and $|\Delta E_\text{mech}| = mgh$
Conversions
We calculate in joules and finally change to kilojoules: $1\,\text{kJ} = 1000\,\text{J}$. The speed is not needed, because the kinetic energy does not change.
Substitution
Answer
$95.6\,\text{kJ}$
Three significant figures like the mass $78.0\,\text{kg}$ and the height $125\,\text{m}$.
Reasonableness
The weight of one kilogram is about $10\,\text{N}$, so $80\,\text{kg}$ and $125\,\text{m}$ give the estimate $800\,\text{N}\cdot 125\,\text{m} = 100\,\text{kJ}$. The kinetic energy $\tfrac12 mv^2 \approx 1\,\text{kJ}$ stays the same, so the whole change in potential energy is changed into other forms of energy. Without air resistance this energy would turn into kinetic energy.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- A sled slides down a straight slope. Why does the normal force of the snow do no work on the sled, although it is a large force?
Answer
The normal force is perpendicular to the slope and so to the motion of the sled. Only the component of the force along the motion does work, $W = F_\parallel s$, and for the normal force it is zero. The normal force does act indirectly, however, because the kinetic friction is $\mu N$.
- When do you choose the equation of motion and when the work–energy principle? Give one example of each from the situations of this chapter.
Answer
The equation of motion $\Sigma\overline{F} = m\overline{a}$ fits when the question is about a force or an acceleration at some moment, for example the reading of a scale in an accelerating lift. The work–energy principle fits when the question is about the speed at some point or the change in energy over a distance, for example the speed of a sled at the foot of a slope. In both you start from the free-body diagram.
- Why does the horizontal distance from the top to the stopping point of a sled starting from rest not depend on the mass of the sled?
Answer
The work $mgh$ done by the weight and the work $-\mu mgx$ done by friction are both proportional to the mass. In the work–energy principle $mgh - \mu mgx = 0$ the mass cancels, and $x = h/\mu$, when the coefficient of friction is the same all the way and air resistance is ignored.
1. Concept
2. Calculation
A student of mass $55.0\,\text{kg}$ stands on a bathroom scale in a lift. The scale measures the normal force and shows it in kilograms, that is, divides it by $9.81\,\text{m/s}^2$. The lift starts downwards and accelerates at $1.20\,\text{m/s}^2$.
What does the scale show? Give the answer in kilograms.
Hint 1: a guiding question
Which is larger, the weight or the normal force, when the acceleration points down?
Hint 2: what is the connection
Choose up as the positive direction: $N - mg = ma$, where $a = -1.20\,\text{m/s}^2$.
Hint 3: an intermediate step
$N = 55.0\,\text{kg}\cdot(9.81 - 1.20)\,\text{m/s}^2$. Finally divide by $9.81\,\text{m/s}^2$.
Show the solution
Given
$m = 55.0\,\text{kg}$, the acceleration of the lift $1.20\,\text{m/s}^2$ downwards, the scale shows the normal force divided by $9.81\,\text{m/s}^2$; $g = 9.81\,\text{m/s}^2$
Find
the reading of the scale $N/g$ in $\text{kg}$
Principle
The student stands still in the lift, so the student's acceleration is the acceleration of the lift. By Newton's second law the acceleration is determined by the sum of the weight and the normal force. The scale measures the normal force.
Free-body diagram
The student as a rectangle, positive direction up. The weight $\overline{G}$ from the centre down and the normal force $\overline{N}$ from the bottom surface up, shorter than the weight, because the acceleration points down. The acceleration down is marked separately and not as a force.
Formula
$N - mg = ma$, so $N = m(g + a)$, and the reading is $\dfrac{N}{g} = \dfrac{m(g + a)}{g}$
Conversions
The acceleration points down, so with up as the positive direction $a = -1.20\,\text{m/s}^2$. No other conversions are needed.
Substitution
Answer
$48.3\,\text{kg}$
Three significant figures like the mass $55.0\,\text{kg}$ and the acceleration $1.20\,\text{m/s}^2$.
Reasonableness
The acceleration is about one eighth of the acceleration of free fall, so the reading decreases by about one eighth: $55\,\text{kg} - 7\,\text{kg} \approx 48\,\text{kg}$. The reading is smaller than the mass, as it should be when the acceleration points down.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
3. Energy at terminal velocity
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 16-1Core
A lift starts upwards from the ground floor and accelerates at $0.80\,\text{m/s}^2$. A person of mass $70.0\,\text{kg}$ stands in the lift. Draw the free-body diagram of the person.
How large is the normal force of the floor on the person?
Answer
$743\,\text{N}$
Show the solution
Given
$m = 70.0\,\text{kg}$, acceleration of the lift $a = 0.80\,\text{m/s}^2$ upwards; $g = 9.81\,\text{m/s}^2$
Find
the normal force $N$ of the floor in $\text{N}$
Principle
The person stands still in the lift, so the person accelerates with the lift. By Newton's second law the acceleration is determined by the net force, so in a lift accelerating upwards the normal force must be larger than the weight.
Free-body diagram
The person as a rectangle, positive direction up. The weight $\overline{G}$ from the centre down and the normal force $\overline{N}$ of the floor from the bottom surface up, longer than the weight. The acceleration up is marked separately and not as a force.
Formula
$N - mg = ma$, so $N = m(g + a)$
Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$N$$= 70.0\,\text{kg}\cdot(9.81 + 0.80)\,\text{m/s}^2$$= 70.0\,\text{kg}\cdot 10.61\,\text{m/s}^2$$= 742.7\,\text{N}$Answer
$743\,\text{N}$
Upwards, three significant figures like the mass $70.0\,\text{kg}$.
Reasonableness
The weight is $70.0\,\text{kg}\cdot 9.81\,\text{m/s}^2 \approx 687\,\text{N}$, and the normal force is $ma = 56\,\text{N}$ larger, about $8\,\%$. The acceleration is about $8\,\%$ of the acceleration of free fall, so the proportion matches.
- Exercise 16-2Core
The mass of a skydiver with equipment is $90.0\,\text{kg}$. When the parachute opens, the air resistance is $2.10\,\text{kN}$ at one moment, and the skydiver is still moving downwards. Draw the free-body diagram of the skydiver.
How large is the net force acting on the skydiver at this moment?
Answer
$1.22\,\text{kN}$
Show the solution
Given
$m = 90.0\,\text{kg}$, air resistance $F_\text{d} = 2.10\,\text{kN}$, the skydiver moves down; $g = 9.81\,\text{m/s}^2$
Find
the net force $\Sigma F$ in $\text{kN}$ and its direction
Principle
The Earth (the weight) and the air (air resistance) act on the skydiver. The net force is the vector sum of the forces. The skydiver moves down, so the air resistance points up, against the weight.
Free-body diagram
The skydiver with the parachute as a rectangle, positive direction up. The weight $\overline{G}$ from the centre down and the air resistance $\overline{F}_\text{d}$ from the top surface up, more than twice as long as the weight.
Formula
$\Sigma F = F_\text{d} - G$, and $G = mg$, so $\Sigma F = F_\text{d} - mg$
Conversions
$2.10\,\text{kN}$$= 2.10\cdot 1000\,\text{N}$$= 2100\,\text{N}$the result is changed back into kilonewtons at the end.
Substitution
$\Sigma F$$= 2100\,\text{N} - 90.0\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 2100\,\text{N} - 882.9\,\text{N}$$= 1217.1\,\text{N}$$\approx 1.22\,\text{kN}$Answer
$1.22\,\text{kN}$
Upwards, three significant figures like the mass $90.0\,\text{kg}$ and the air resistance $2.10\,\text{kN}$.
Reasonableness
The air resistance is about $2.4$ times the weight $883\,\text{N}$, so about $1.4$ weights are left upwards, about $1.2\,\text{kN}$. The net force points up although the skydiver moves down, so the skydiver slows down.
- Exercise 16-3
A child slides down an icy chute from rest. The start of the chute is $3.20\,\text{m}$ higher than its end, and friction and air resistance can be ignored. Draw the free-body diagram of the child.
How large is the speed of the child at the end of the chute?
Answer
$7.92\,\text{m/s}$
Show the solution
Given
height difference $h = 3.20\,\text{m}$, initial speed $v_1 = 0$, friction and air resistance are ignored; $g = 9.81\,\text{m/s}^2$
Find
the speed $v_2$ at the end of the chute in $\text{m/s}$
Free-body diagram
The child as a rectangle, tilted along the surface of the chute. The weight $\overline{G}$ from the centre straight down and the normal force $\overline{N}$ of the chute from the bottom surface perpendicularly out of the surface. There is no friction and no force in the direction of the motion. The velocity is marked separately and not as a force.
Principle
The weight and the normal force of the chute act on the child. The normal force is perpendicular to the motion and does no work, and friction is ignored, so the mechanical energy is conserved. The shape of the chute does not matter, only the height difference.
Formula
Zero level at the end of the chute: $mgh = \tfrac12 mv_2^2$, so $v_2 = \sqrt{2gh}$
Conversions
The quantities are in SI units, so no conversions are needed. The mass is not needed, because it cancels.
Substitution
$v_2$$= \sqrt{2\cdot 9.81\,\text{m/s}^2\cdot 3.20\,\text{m}}$$= \sqrt{62.78\,\text{m}^2/\text{s}^2}$$= 7.924\,\text{m/s}$Answer
$7.92\,\text{m/s}$
Three significant figures like the height $3.20\,\text{m}$.
Reasonableness
$7.9\,\text{m/s}$ is about $29\,\text{km/h}$, which suits an icy chute. In reality friction and air resistance reduce the speed a little, so the result is an upper limit.
- Exercise 16-4
A sled arrives from the foot of a slope onto a horizontal snow field at a speed of $6.0\,\text{m/s}$. The coefficient of kinetic friction between the sled and the snow is $0.080$, and air resistance is ignored.
How long a distance does the sled slide on the flat before it stops?
Answer
$23\,\text{m}$
Show the solution
Given
initial speed $v_1 = 6.0\,\text{m/s}$, final speed $v_2 = 0$, $\mu = 0.080$; $g = 9.81\,\text{m/s}^2$
Find
the sliding distance $s$ in $\text{m}$
Principle
The question is about the distance over which the kinetic energy is used up, so we use the work–energy principle. The height does not change, and friction is the only force doing work: its negative work turns the kinetic energy into internal energy. The friction force comes from the free-body diagram.
Free-body diagram
The sled as a rectangle. The weight $\overline{G}$ from the centre down, the normal force $\overline{N}$ of the snow from the bottom surface up, equally long, and the kinetic friction $\overline{F}_\mu$ from the rear corner of the bottom surface backwards. There is no force pointing forwards. In the vertical direction $N - mg = 0$, so $F_\mu = \mu mg$.
Formula
$\tfrac12 mv_1^2 - \mu mg s = 0$, so $s = \dfrac{v_1^2}{2\mu g}$
Conversions
The quantities are in SI units, so no conversions are needed. The mass cancels.
Substitution
$s$$= \dfrac{(6.0\,\text{m/s})^2}{2\cdot 0.080\cdot 9.81\,\text{m/s}^2}$$= \dfrac{36\,\text{m}^2/\text{s}^2}{1.5696\,\text{m/s}^2}$$= 22.94\,\text{m}$Answer
$23\,\text{m}$
Two significant figures like the speed $6.0\,\text{m/s}$ and the coefficient of friction $0.080$.
Reasonableness
The deceleration is $\mu g \approx 0.8\,\text{m/s}^2$, so stopping takes about $6\,\text{s} : 0.8 \approx 7.5\,\text{s}$, and at the average speed $3\,\text{m/s}$ the distance is about $23\,\text{m}$. On the flat of a sledding hill this is a usual sliding distance.
- Exercise 16-5
A sled slides down a straight slope at a steady speed. The coefficient of kinetic friction between the sled and the snow is $0.15$, and air resistance is ignored.
How large is the angle of inclination of the slope?
Answer
$8.5^\circ$
Show the solution
Given
$\mu = 0.15$, the speed is constant, air resistance is ignored
Find
the angle of inclination $\alpha$ of the slope in degrees
Principle
The speed is constant, so by Newton's second law the net force is zero both along the slope and perpendicular to it. The component of the weight along the slope and the kinetic friction cancel each other.
Free-body diagram
The sled as a rectangle along the slope, $x$ down the slope and $y$ perpendicularly out of the slope. The weight $\overline{G}$ from the centre straight down, the normal force $\overline{N}$ from the bottom surface perpendicularly out of the slope and the kinetic friction $\overline{F}_\mu$ from the upper corner of the bottom surface up the slope. The components of the weight as dashed lines.
Formula
$y$: $N - mg\cos\alpha = 0$; $x$: $mg\sin\alpha - \mu N = 0$, so $mg\sin\alpha = \mu mg\cos\alpha$, $\tan\alpha = \mu$ and $\alpha = \tan^{-1}\mu$
Conversions
The coefficient of friction has no unit, and the angle comes in degrees with the calculator in degree mode, so no conversions are needed.
Substitution
$\alpha = \tan^{-1} 0.15 = 8.531^\circ$
Answer
$8.5^\circ$
Two significant figures like the coefficient of friction $0.15$.
Reasonableness
For small angles $\tan\alpha \approx \alpha$ in radians, and $0.15\,\text{rad} \approx 8.6^\circ$. The angle is gentle, which fits: on a steeper slope the sled would accelerate, on a gentler one it would slow down. The mass cancelled.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 16-6Core
A student of mass $60.0\,\text{kg}$ stands on a bathroom scale in a lift. The scale measures the normal force and shows it in kilograms, that is, divides it by $9.81\,\text{m/s}^2$. For a moment the scale shows $66.5\,\text{kg}$.
How large is the acceleration of the lift at this moment, and in which direction does it point?
Answer
$1.06\,\text{m/s}^2$
Show the solution
Given
$m = 60.0\,\text{kg}$, reading of the scale $\tfrac{N}{g} = 66.5\,\text{kg}$; $g = 9.81\,\text{m/s}^2$
Find
the acceleration $a$ of the lift in $\text{m/s}^2$ and its direction
Principle
A scale measures the normal force. The reading is larger than the mass, so the normal force is larger than the weight, the net force points up and by Newton's second law the acceleration also points up.
Free-body diagram
The student as a rectangle, positive direction up. The weight $\overline{G}$ from the centre down and the normal force $\overline{N}$ from the bottom surface up, longer than the weight.
Formula
$N - mg = ma$, so $a = \dfrac{N - mg}{m}$. When the reading is $L = \tfrac{N}{g}$, $N = Lg$ and $a = g\left(\dfrac{L}{m} - 1\right)$
Conversions
The reading is changed into the normal force by multiplying by $g$:
$N$$= 66.5\,\text{kg}\cdot 9.81\,\text{m/s}^2$$= 652.4\,\text{N}$Substitution
$a$$= 9.81\,\text{m/s}^2\cdot\left(\dfrac{66.5\,\text{kg}}{60.0\,\text{kg}} - 1\right)$$= 9.81\,\text{m/s}^2\cdot 0.10833$$= 1.063\,\text{m/s}^2$Answer
$1.06\,\text{m/s}^2$
Upwards, three significant figures like the mass $60.0\,\text{kg}$ and the reading $66.5\,\text{kg}$.
Reasonableness
The reading is about $11\,\%$ larger than the mass, so the acceleration is about $11\,\%$ of the acceleration of free fall, about $1\,\text{m/s}^2$. The lift either starts up or slows down while moving down: both give an acceleration pointing up.
- Exercise 16-7Core
A sled starts from rest at the top of a straight slope. The slope is $18.0\,\text{m}$ long and at an angle of $12.0^\circ$ to the horizontal, and at the foot of the slope the speed of the sled is $5.10\,\text{m/s}$. Air resistance is ignored.
How large is the coefficient of kinetic friction between the sled and the snow?
Answer
$0.137$
Show the solution
Given
$L = 18.0\,\text{m}$, $\alpha = 12.0^\circ$, $v_1 = 0$, $v_2 = 5.10\,\text{m/s}$; $g = 9.81\,\text{m/s}^2$
Find
the coefficient of kinetic friction $\mu$ (no unit)
Principle
The speed is known at the start and at the end of the slope, so we use the work–energy principle. The weight does work through the potential energy, the normal force does no work, and kinetic friction does negative work along the whole slope. The friction force comes from the free-body diagram.
Free-body diagram
The sled as a rectangle along the slope, $x$ down the slope and $y$ perpendicularly out of the slope. The weight $\overline{G}$ from the centre straight down, the normal force $\overline{N}$ from the bottom surface perpendicularly out of the slope and the kinetic friction $\overline{F}_\mu$ from the upper corner of the bottom surface up the slope. In the $y$ direction $N = mg\cos\alpha$, so $F_\mu = \mu mg\cos\alpha$.
Formula
Zero level at the foot: $mgL\sin\alpha - \mu mg\cos\alpha\cdot L = \tfrac12 mv_2^2$, so
$\mu$$= \dfrac{gL\sin\alpha - \tfrac12 v_2^2}{gL\cos\alpha}$Conversions
The quantities are in SI units, so no conversions are needed. The mass cancels.
Substitution
The numerator is
$9.81\cdot 18.0\cdot\sin 12.0^\circ\,\tfrac{\text{m}^2}{\text{s}^2} - \tfrac12\cdot 5.10^2\,\tfrac{\text{m}^2}{\text{s}^2}$$= 36.71\,\tfrac{\text{m}^2}{\text{s}^2} - 13.01\,\tfrac{\text{m}^2}{\text{s}^2}$$= 23.71\,\tfrac{\text{m}^2}{\text{s}^2}$and the denominator
$9.81\cdot 18.0\cdot\cos 12.0^\circ\,\tfrac{\text{m}^2}{\text{s}^2}$$= 172.72\,\tfrac{\text{m}^2}{\text{s}^2}$so $\mu = \dfrac{23.71}{172.72} = 0.1373$
Answer
$0.137$
Three significant figures like the length of the slope $18.0\,\text{m}$, the angle $12.0^\circ$ and the speed $5.10\,\text{m/s}$.
Reasonableness
Without friction the speed would be $\sqrt{2\cdot 9.81\cdot 3.74}\,\text{m/s} \approx 8.6\,\text{m/s}$ from the drop height
$h$$= 18.0\,\text{m}\cdot\sin 12.0^\circ$$\approx 3.74\,\text{m}$so friction takes about two thirds of the energy. The coefficient is smaller than $\tan 12.0^\circ \approx 0.21$, as it should be, because the sled accelerates.
- Exercise 16-8
A skydiver whose mass with equipment is $80.0\,\text{kg}$ falls at a speed of $52.0\,\text{m/s}$ and opens the parachute. While falling $145\,\text{m}$, the skydiver's speed decreases to $6.0\,\text{m/s}$. Draw the free-body diagram of the skydiver.
How much mechanical energy is changed into other forms of energy over this distance?
Answer
$221\,\text{kJ}$
Show the solution
Given
$m = 80.0\,\text{kg}$, $v_1 = 52.0\,\text{m/s}$, $v_2 = 6.0\,\text{m/s}$, drop $h = 145\,\text{m}$; $g = 9.81\,\text{m/s}^2$
Find
the mechanical energy changed into other forms of energy $|\Delta E_\text{mech}|$ in $\text{kJ}$
Free-body diagram
The skydiver with the parachute as a rectangle, positive direction down. The weight $\overline{G}$ from the centre down and the air resistance $\overline{F}_\text{d}$ from the top surface up. The velocity down is marked separately and not as a force. The air resistance is against the motion, so the work it does is negative.
Principle
By the work–energy principle the change in mechanical energy is the work done by forces other than the weight, here the negative work done by air resistance. Both the kinetic energy and the potential energy decrease, and the change in both is counted.
Formula
Zero level at the height of the final situation: $\Delta E_\text{mech} = \left(\tfrac12 mv_2^2 - \tfrac12 mv_1^2\right) - mgh$, and
$|\Delta E_\text{mech}|$$= \dfrac{m\left(v_1^2 - v_2^2\right)}{2} + mgh$Conversions
We calculate in joules and finally change to kilojoules: $1\,\text{kJ} = 1000\,\text{J}$.
Substitution
$|\Delta E_\text{mech}|$$= \tfrac12\cdot 80.0\,\text{kg}\cdot\left(52.0^2 - 6.0^2\right)\,\text{m}^2/\text{s}^2 + 80.0\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot 145\,\text{m}$$= 106\,720\,\text{J} + 113\,796\,\text{J}$$= 220\,516\,\text{J}$$\approx 221\,\text{kJ}$Answer
$221\,\text{kJ}$
Three significant figures like the mass $80.0\,\text{kg}$ and the speed $52.0\,\text{m/s}$.
Reasonableness
Each part is about $110\,\text{kJ}$: the kinetic energy decreases almost completely ($\tfrac12\cdot 80\cdot 52^2\,\text{J} \approx 108\,\text{kJ}$), and the potential energy decreases by
$mgh$$\approx 800\,\text{N}\cdot 145\,\text{m}$$\approx 116\,\text{kJ}$The energy does not disappear but turns into motion of the air and finally into internal energy.
- Exercise 16-9
The combined mass of a lift car and its load is $850\,\text{kg}$. A motor raises the car with a cable at a steady speed $12.0\,\text{m}$ upwards, and the lift takes $15.0\,\text{s}$.
How large is the power the cable gives to the car?
Answer
$6.67\,\text{kW}$
Show the solution
Given
$m = 850\,\text{kg}$, lifting height $h = 12.0\,\text{m}$, time $t = 15.0\,\text{s}$, the speed is constant; $g = 9.81\,\text{m/s}^2$
Find
the power $P$ of the cable in $\text{kW}$
Principle
The speed is constant, so by Newton's second law the tension of the cable is equal to the weight. The cable pulls in the direction of the motion, so it does positive work $W = Th$, which increases the potential energy of the car. Power is work divided by time, as in the chapter Hydro and wind power as energy converters, $E = Pt$.
Free-body diagram
The lift car with its load as a rectangle, positive direction up. The weight $\overline{G}$ from the centre down and the tension $\overline{T}$ of the cable from the top surface up, equally long. The velocity up is marked separately and not as a force.
Formula
$T - mg = 0$, so $T = mg$. The work is $W = Th = mgh$, and the power $P = \dfrac{W}{t} = \dfrac{mgh}{t}$
Conversions
We calculate in watts and finally change to kilowatts: $1\,\text{kW} = 1000\,\text{W}$.
Substitution
$P$$= \dfrac{850\,\text{kg}\cdot 9.81\,\text{m/s}^2\cdot 12.0\,\text{m}}{15.0\,\text{s}}$$= \dfrac{100\,062\,\text{J}}{15.0\,\text{s}}$$= 6670.8\,\text{W}$$\approx 6.67\,\text{kW}$Answer
$6.67\,\text{kW}$
Three significant figures like the height $12.0\,\text{m}$ and the time $15.0\,\text{s}$.
Reasonableness
The speed is $12.0\,\text{m} : 15.0\,\text{s} = 0.80\,\text{m/s}$, and the tension is about $8.3\,\text{kN}$. The product $8.3\,\text{kN}\cdot 0.80\,\text{m/s} \approx 6.7\,\text{kW}$ matches. The power of a lift motor is typically a few kilowatts, and because of losses the motor needs more.
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 16-10
The top of a sledding hill is $4.50\,\text{m}$ higher than the flat snow field that begins at the foot of the hill. The slope of the hill consists of two straight parts: the upper one is at an angle of $25^\circ$ and the lower one at $10^\circ$ to the horizontal.
The sled starts from rest at the top, and the coefficient of kinetic friction is $0.120$ all the way. Air resistance and the increase of friction in the bends of the slope are ignored.
How far from the top, measured horizontally, does the sled stop?
Answer
$37.5\,\text{m}$
Show the solution
Given
height difference $h = 4.50\,\text{m}$, $\mu = 0.120$ all the way, parts of the slope $25^\circ$ and $10^\circ$, $v_1 = 0$ and at the end $v_2 = 0$
Find
the horizontal distance $x$ from the top to the stopping point in $\text{m}$
Principle
The sled is at rest at the start and at the end, so by the work–energy principle the negative work done by friction is equal to the decrease in potential energy. On a straight surface the work done by friction is $-\mu mg$ times the horizontal distance, because on a slope $N = mg\cos\alpha$ and $L\cos\alpha$ is the horizontal length of the slope. So the angles of the parts of the slope are not needed.
Free-body diagram
On each part of the slope the sled as a rectangle along the slope: the weight $\overline{G}$ from the centre down, the normal force $\overline{N}$ perpendicularly out of the slope, $N = mg\cos\alpha$, and the kinetic friction $\overline{F}_\mu$ up the slope. On the flat $N = mg$ and the friction points backwards. The normal force does no work anywhere.
Formula
Horizontal distances of the parts $x_\text{a}$, $x_\text{b}$ and $x_\text{c}$: $mgh - \mu mg(x_\text{a} + x_\text{b} + x_\text{c}) = 0$, so
$x$$= x_\text{a} + x_\text{b} + x_\text{c}$$= \dfrac{h}{\mu}$Conversions
The quantities are in SI units and the coefficient of friction has no unit, so no conversions are needed.
Substitution
$x$$= \dfrac{4.50\,\text{m}}{0.120}$$= 37.5\,\text{m}$Answer
$37.5\,\text{m}$
Three significant figures like the height $4.50\,\text{m}$ and the coefficient of friction $0.120$.
Reasonableness
The inclination of both parts of the slope is greater than $\tan^{-1} 0.120 \approx 6.8^\circ$, so the sled slides down the whole slope and does not stop halfway. The result does not depend on the mass or on the shape of the slope. A smaller friction would give a longer distance, as it should. In reality the normal force and the friction grow for a moment in the bends, so the sled would stop a little earlier.
- Exercise 16-11
A person of mass $75.0\,\text{kg}$ stands on a force-sensor plate in a lift that starts downwards from an upper floor and stops at a lower one. The graph shows the reading of the plate, that is, the normal force, as a function of time.
How large is the greatest speed of the lift during the trip?
Answer
$2.3\,\text{m/s}$
Show the solution
Given
$m = 75.0\,\text{kg}$; from the graph the normal force $N = 650\,\text{N}$ from $1.0\,\text{s}$ to $3.0\,\text{s}$; the lift starts from rest; $g = 9.81\,\text{m/s}^2$
Find
the greatest speed $v$ of the lift in $\text{m/s}$
Principle
The lift accelerates downwards when the normal force is smaller than the weight, and the speed is greatest at the end of this phase. By Newton's second law the acceleration comes from the forces. When the forces are constant, the acceleration is constant, and by the chapter Acceleration and uniformly accelerated motion $v = v_0 + at$.
Free-body diagram
The person as a rectangle, positive direction down. The weight $\overline{G}$ from the centre down and the normal force $\overline{N}$ from the bottom surface up, shorter than the weight.
Reading the graph
The weight is $75.0\,\text{kg}\cdot 9.81\,\text{m/s}^2 = 735.75\,\text{N}$, and the reading is that large at rest and at a steady speed. From $1.0\,\text{s}$ to $3.0\,\text{s}$ the curve is on the grid line at $650\,\text{N}$, smaller than the weight, so the lift accelerates downwards for $\Delta t = 2.0\,\text{s}$. From $6.0\,\text{s}$ on the reading is larger than the weight, so the lift slows down.
Formula
$mg - N = ma$, so $a = g - \dfrac{N}{m}$, and starting from rest
$v$$= a\Delta t$$= \left(g - \dfrac{N}{m}\right)\Delta t$Conversions
The quantities are in SI units, so no conversions are needed.
Substitution
$v$$= \left(9.81\,\text{m/s}^2 - \dfrac{650\,\text{N}}{75.0\,\text{kg}}\right)\cdot 2.0\,\text{s}$$= 1.1433\,\text{m/s}^2\cdot 2.0\,\text{s}$$= 2.287\,\text{m/s}$Answer
$2.3\,\text{m/s}$
Downwards, two significant figures like the time $2.0\,\text{s}$ read from the graph.
Reasonableness
In the slowing phase the reading is $800\,\text{N}$, so the deceleration is $(800 - 735.75)\,\text{N} : 75.0\,\text{kg} \approx 0.86\,\text{m/s}^2$, and the speed $2.3\,\text{m/s}$ disappears in about $2.7\,\text{s}$, which matches the slowing phase of the graph from about $6.0\,\text{s}$ to $8.7\,\text{s}$. The speed of an ordinary lift is $1$–$3\,\text{m/s}$.
- Exercise 16-12
A skydiver whose mass with equipment is $70.0\,\text{kg}$ opens the parachute at a speed of $50.0\,\text{m/s}$. After falling $60.0\,\text{m}$ the speed has decreased to $6.0\,\text{m/s}$.
How large is the air resistance on average over this distance, if it is assumed to be constant?
Answer
$2.12\,\text{kN}$
Show the solution
Given
$m = 70.0\,\text{kg}$, $v_1 = 50.0\,\text{m/s}$, $v_2 = 6.0\,\text{m/s}$, drop $h = 60.0\,\text{m}$, air resistance assumed constant; $g = 9.81\,\text{m/s}^2$
Find
the magnitude of the air resistance $F_\text{d}$ in $\text{kN}$
Principle
The speeds are known at the start and at the end of the distance, so we use the work–energy principle. Air resistance is the only force other than the weight, and it points against the motion, so the work it does is $W_\text{d} = -F_\text{d} h$. The work comes from the work–energy principle, and the force from the work.
Free-body diagram
The skydiver with the parachute as a rectangle. The weight $\overline{G}$ from the centre down and the air resistance $\overline{F}_\text{d}$ from the top surface up, much longer than the weight, because the skydiver slows down. The velocity down is marked separately and not as a force.
Formula
The kinetic energies are $E_{\text{k},1} = \tfrac12 mv_1^2$ and $E_{\text{k},2} = \tfrac12 mv_2^2$. Zero level at the height of the final situation: $E_{\text{k},1} + mgh - F_\text{d} h = E_{\text{k},2}$, so
$F_\text{d}$$= mg + \dfrac{E_{\text{k},1} - E_{\text{k},2}}{h}$Conversions
We calculate in newtons and finally change to kilonewtons: $1\,\text{kN} = 1000\,\text{N}$.
Substitution
$E_{\text{k},1}$$= \tfrac12\cdot 70.0\,\text{kg}\cdot(50.0\,\text{m/s})^2$$= 87\,500\,\text{J}$$E_{\text{k},2}$$= \tfrac12\cdot 70.0\,\text{kg}\cdot(6.0\,\text{m/s})^2$$= 1260\,\text{J}$so
$F_\text{d}$$= 70.0\,\text{kg}\cdot 9.81\,\text{m/s}^2 + \dfrac{87\,500\,\text{J} - 1260\,\text{J}}{60.0\,\text{m}}$$= 686.7\,\text{N} + 1437.3\,\text{N}$$= 2124\,\text{N}$$\approx 2.12\,\text{kN}$Answer
$2.12\,\text{kN}$
Upwards, three significant figures like the mass $70.0\,\text{kg}$, the speed $50.0\,\text{m/s}$ and the distance $60.0\,\text{m}$.
Reasonableness
The air resistance is about three times the weight $687\,\text{N}$, so the net force points up and the skydiver slows down, as it should. From the equation of motion the deceleration is $(2124 - 687)\,\text{N} : 70.0\,\text{kg} \approx 20\,\text{m/s}^2$, and then
$\tfrac12(v_1^2 - v_2^2) : a$$\approx 1232 : 20.5$$\approx 60\,\text{m}$as in the problem.