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Energy · Work and power

Work

A musher pulls a sled with the rope at an angle; a worker pushes a pallet in a straight line; a gondola lift in Colorado climbs and then descends part of the way. In all three cases there are forces and there is displacement, but the work done depends on how the two vectors combine. In this page you will calculate work using the angle or using components, and you will use the work–energy theorem to predict speeds without needing to know forces at each instant or times.

Work done by a constant force: $W = F\,d\cos\theta$

You push with all your strength against a concrete wall for ten minutes and end up exhausted, but the wall does not move; on the other hand, with a gentle push you send a skateboard rolling several meters. Why does physics say that in the first case you have done no mechanical work on the wall and in the second you have?

Three limiting cases are worth memorizing. At $\theta = 0^\circ$ the force points along the displacement and the work is positive and largest ($W = Fd$). At $\theta = 90^\circ$ the force is perpendicular to the displacement and the work is zero. At $\theta = 180^\circ$ the force points against the displacement and the work is negative ($W = -Fd$).

Of the whole force, only the component in the direction of the displacement, $F_{\parallel} = F \cos \theta$, "counts" toward the work. The perpendicular component, $F \sin \theta$, contributes no work.

Work can also be read from a graph. If you plot $F_{\parallel}$ against position $x$, the work is the area between the graph and the $x$-axis. For a constant force that area is a rectangle of height $F_{\parallel}$ and width $d$, which gives $W = F_{\parallel}\, d$. For a force that changes, the area is no longer a rectangle. A spring pulled from $x=0$ to $x$ has $F_s = kx$, so the graph is a straight line from $0$ to $kx$ and the area is a triangle: $W = \frac12 \cdot x \cdot kx = \frac12 k x^2$. For example, with $k = 200\ \text{N/m}$ stretched $0.10\ \text{m}$, $W = \frac12 \cdot 0.10 \cdot 20 = 1.0\ \text{J}$.

Try the widget below: change the angle of the force and watch how the component along the horizontal displacement, $F\cos\theta$, changes.

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The work done by a constant force on a moving body is $W = \vec{F} \cdot \vec{d} = F \cdot d \cdot \cos \alpha$, where $\alpha$ is the angle between the force and the…

The work done by a constant force on a moving body is $W = \vec{F} \cdot \vec{d} = F \cdot d \cdot \cos \alpha$, where $\alpha$ is the angle between the force and the displacement. Without performing calculations, state whether the work is positive, negative, or zero in each case, justifying your answer using the value of $\alpha$:

  1. a) The pulling force of a tractor towing a trailer along a horizontal path.
  2. b) The friction force opposing the motion of the trailer.
  3. c) The normal force exerted by the ground on the trailer.
  4. d) The force with which a worker holds a box at rest on their shoulder.
Solve
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A warehouse worker pushes a pallet with a constant horizontal force of $250\ \text{N}$, moving it $12\ \text{m}$ along the floor in a straight line. a) What angle does…

A warehouse worker pushes a pallet with a constant horizontal force of $250\ \text{N}$, moving it $12\ \text{m}$ along the floor in a straight line.

  1. a) What angle does the force make with the displacement?
  2. b) Calculate the work done by the worker.
Solve
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A person walks their dog, pulling on the leash with a constant force of $80\ \text{N}$ at an angle of $60^\circ$ to the horizontal over a straight section of…

A person walks their dog, pulling on the leash with a constant force of $80\ \text{N}$ at an angle of $60^\circ$ to the horizontal over a straight section of $25\ \text{m}$ ($\cos 60^\circ = 0.5$), as shown in the figure.

F = 80 N 60° d = 25 m W = ?

Calculate the work done.

Solve
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A traveler pushes their luggage cart through an airport terminal in a straight line on a horizontal floor. a) Reason, using the expression…

A traveler pushes their luggage cart through an airport terminal in a straight line on a horizontal floor.

  1. a) Reason, using the expression $W = F \times d \times \cos \alpha$, why the work done by the normal force and the work done by the weight are zero during the trip.
  2. b) If the traveler exerts a horizontal force of $18\ \text{N}$ over $100\ \text{m}$, what work do they do?
Solve
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A musher pulls their team's sled using a rope at an angle of $37^\circ$ to the horizontal, exerting a constant force of $100\ \text{N}$ over $50\ \text{m}$ (…

A musher pulls their team's sled using a rope at an angle of $37^\circ$ to the horizontal, exerting a constant force of $100\ \text{N}$ over $50\ \text{m}$ ($\sen 37^\circ \approx 0.6$; $\cos 37^\circ \approx 0.8$). Friction with the snow opposes with $30\ \text{N}$.

  1. a) Calculate the work done by the musher.
  2. b) Calculate the work done by friction.
  3. c) Find the net work done on the sled.
Solve

Work in components: $W = F_x \Delta x + F_y \Delta y$

In a space-navigation video game, the thrusters apply a force with components $F_x = 12\ \text{N}$ and $F_y = -5\ \text{N}$ to a probe while it moves from $(0, 0)$ to $(8, 3)\ \text{m}$, and measuring the angle between the force and the displacement is not practical. How can we calculate the energy transferred to the ship by using components directly?

A force resolved into horizontal and vertical components, showing the horizontal component Fx in the same direction as the displacement displacement Δx F (Fx, Fy) Fx
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In the lab, a cart on an air track moves $\vec{d} = (5, 0)\ \text{m}$ while acted upon by a constant force $\vec{F} = (6, 8)\ \text{N}$, as shown in the figure. a)…

In the lab, a cart on an air track moves $\vec{d} = (5, 0)\ \text{m}$ while acted upon by a constant force $\vec{F} = (6, 8)\ \text{N}$, as shown in the figure.

x y d = (5, 0) m F = (6, 8) N Fₓ Fy
  1. a) Calculate the work using the dot product $W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y$.
  2. b) Explain why the vertical component of the force does not contribute to the work.
Solve
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A constant force $\vec{F} = (15, -8)\ \text{N}$ acts on a machine component while the component moves through a displacement of $\vec{d} = (4, 2)\ \text{m}$, as shown in…

A constant force $\vec{F} = (15, -8)\ \text{N}$ acts on a machine component while the component moves through a displacement of $\vec{d} = (4, 2)\ \text{m}$, as shown in the figure.

x y d = (4, 2) m F = (15, −8) N Fₓ Fy
  1. a) Calculate the work using the dot product $W = \vec{F} \cdot \vec{d} = F_x \cdot d_x + F_y \cdot d_y$.
  2. b) Check that the component $F_y$ does negative work and interpret what that sign means physically.
Solve
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On a laboratory air table, a puck of $4\ \text{kg}$, initially at rest at point $A(1, 2)$, moves to point $B(9, 5)$ while a constant force $\vec{F} = (6, 8)\ \text{N}$…

On a laboratory air table, a puck of $4\ \text{kg}$, initially at rest at point $A(1, 2)$, moves to point $B(9, 5)$ while a constant force $\vec{F} = (6, 8)\ \text{N}$ acts on it (the coordinates are in meters), as shown in the figure.

x y 2 4 6 8 2 4 A(1, 2) B(9, 5) F = (6, 8) N
  1. a) Determine the displacement vector $\vec{d} = \vec{AB}$.
  2. b) Calculate the work done by the force using the scalar product $W = \vec{F} \cdot \vec{d}$.
  3. c) Since friction is negligible, this work is the net work: apply the work-energy theorem and calculate the speed of the puck when it reaches $B$.
Solve

Friction and net work on an inclined plane

Free-body diagram of a block moving up an inclined plane: pulling force and friction parallel to the ramp, normal force perpendicular to it, and vertical weight θ FT pull friction FN (normal) Fg
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When dragging a sofa $8\ \text{m}$ across a living room floor, the kinetic friction force is $35\ \text{N}$. a) What angle does the friction force make with the…

When dragging a sofa $8\ \text{m}$ across a living room floor, the kinetic friction force is $35\ \text{N}$.

  1. a) What angle does the friction force make with the displacement?
  2. b) Calculate the work done by friction and interpret its sign.
Solve
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In a scavenger hunt, a student drags a box of books $15\ \text{m}$ across the horizontal floor of the gymnasium by pushing with a horizontal force of $120\ \text{N}$,…

In a scavenger hunt, a student drags a box of books $15\ \text{m}$ across the horizontal floor of the gymnasium by pushing with a horizontal force of $120\ \text{N}$, while kinetic friction opposes the motion with $40\ \text{N}$, as shown in the figure.

box F = 120 N friction = 40 N d = 15 m W = ?
  1. a) Calculate the work done by the student.
  2. b) Calculate the work done by friction (remember it is negative: $\alpha = 180^\circ$).
  3. c) Calculate the work done by weight and by the normal force, justifying their values.
  4. d) Find the net work done on the box.
Solve
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To load a refrigerator of $60\ \text{kg}$ into a van, it is pushed up a ramp of $3\ \text{m}$ at an angle of $20^\circ$ to the horizontal ($\sen 20^\circ \approx 0.34$;…

To load a refrigerator of $60\ \text{kg}$ into a van, it is pushed up a ramp of $3\ \text{m}$ at an angle of $20^\circ$ to the horizontal ($\sen 20^\circ \approx 0.34$; take $g = 10\ \text{m/s}^2$), as shown in the figure.

20° 60 kg P pₓ = ? 3 m
  1. a) Calculate the component of the weight parallel to the ramp.
  2. b) Calculate the work done by the weight during the ascent.
  3. c) What minimum work must the person pushing do against that component of weight?
Solve
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A box of $25\ \text{kg}$ starts from rest on the horizontal floor of a warehouse, where the coefficient of kinetic friction is $\mu = 0.2$. It is pushed with a constant…

A box of $25\ \text{kg}$ starts from rest on the horizontal floor of a warehouse, where the coefficient of kinetic friction is $\mu = 0.2$. It is pushed with a constant horizontal force of $100\ \text{N}$ over $12\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.

μ = 0.2 25 kg F = 100 N Froz = ? N P d = 12 m
  1. a) Calculate the kinetic friction force, $F_{roz} = \mu \times N$, taking into account that on a horizontal floor the normal force is $N = m \times g$.
  2. b) Calculate the work done by each of the forces (push, friction, weight, and normal force) and the net work.
  3. c) Apply the work-energy theorem to find the speed of the box at the end of the $12\ \text{m}$.
Solve
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A worker pulls a $25\ \text{kg}$ box up a $20\ \text{m}$ ramp inclined at $37^\circ$ to the horizontal, pulling with a rope parallel to the ramp and at constant speed…

A worker pulls a $25\ \text{kg}$ box up a $20\ \text{m}$ ramp inclined at $37^\circ$ to the horizontal, pulling with a rope parallel to the ramp and at constant speed ($\sen 37^\circ \approx 0.6$; $\cos 37^\circ \approx 0.8$; coefficient of friction $\mu = 0.25$; take $g = 10\ \text{m/s}^2$), as shown in the figure.

37° μ = 0.25 25 kg F = ? friction = ? N = ? P 20 m
  1. a) Calculate the normal force exerted by the ramp and the friction force.
  2. b) Calculate the force with which the worker pulls.
  3. c) Calculate the work done by each force (pull, weight, friction, and normal force) and check that the net work is zero. Why did it have to be zero?
Solve

The work–energy theorem: $W_{net} = \Delta K$

Road-safety tests show that a car braking hard at $10\ \text{m/s}$ stops in about $10\ \text{m}$, but at twice the speed ($20\ \text{m/s}$) it does not need twice the distance, but about $40\ \text{m}$. What physical principle links the work done by the brakes with the loss of speed of the vehicle without needing to calculate how long the braking lasts?

Why? For a constant net force, Newton's second law gives $F_{net} = ma$, and the kinematics equation $v^2 = v_0^2 + 2ad$ gives $ad = \frac12(v^2 - v_0^2)$. So $W_{net} = F_{net} d = m a d = \frac12 m v^2 - \frac12 m v_0^2$.

To use speeds in $\text{m/s}$, convert highway speeds first: $1\ \text{mph} \approx 0.447\ \text{m/s}$, so $72\ \text{mph} \approx 32\ \text{m/s}$ and $60\ \text{mph} \approx 27\ \text{m/s}$. (The speed in km/h used in some problems converts with $1\ \text{km/h} = \frac{1}{3.6}\ \text{m/s}$.)

A car reduces its speed from v₀ to v over a distance d under a net force that opposes the motion; the net work of that force is equal to the decrease in kinetic energy, shown with two bars v₀ v F d ½mv₀² ½mv² K initial K final ΔK W net = ΔK W net: work done by the net force ΔK: change in kinetic energy
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The work-energy theorem states that the net work done on a body is equal to the change in its kinetic energy:…

The work-energy theorem states that the net work done on a body is equal to the change in its kinetic energy: $W_{neto} = \Delta E_c = \frac{1}{2} \times m \times v^2 - \frac{1}{2} \times m \times v_0^2$. A demonstration ball of $2\ \text{kg}$ increases its speed from $3\ \text{m/s}$ to $5\ \text{m/s}$ on an air track.

  1. a) Calculate its initial kinetic energy and its final kinetic energy.
  2. b) What is the net work done on the ball?
Solve
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A passenger car of $1200\ \text{kg}$ accelerates from rest to $108\ \text{km/h}$ when merging onto a divided highway. a) Express the final velocity in $\text{m/s}$. b)…

A passenger car of $1200\ \text{kg}$ accelerates from rest to $108\ \text{km/h}$ when merging onto a divided highway.

  1. a) Express the final velocity in $\text{m/s}$.
  2. b) Calculate the change in kinetic energy of the car.
  3. c) What net work has been done by the forces acting on the car? Express it in $\text{kJ}$.
Solve
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A car of $1000\ \text{kg}$ traveling at $20\ \text{m/s}$ brakes to a stop over a distance of $50\ \text{m}$ along a horizontal section of road. a) Calculate the change…

A car of $1000\ \text{kg}$ traveling at $20\ \text{m/s}$ brakes to a stop over a distance of $50\ \text{m}$ along a horizontal section of road.

  1. a) Calculate the change in kinetic energy during braking.
  2. b) Applying the work-energy theorem, find the work done by the net braking force.
  3. c) What is the average value of this net braking force? State its direction.
Solve
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A passenger car of $1000\ \text{kg}$ travels at $20\ \text{m/s}$ along a horizontal road and locks its wheels during emergency braking. The coefficient of friction…

A passenger car of $1000\ \text{kg}$ travels at $20\ \text{m/s}$ along a horizontal road and locks its wheels during emergency braking. The coefficient of friction between the tires and the asphalt is $\mu = 0.5$ (take $g = 10\ \text{m/s}^2$).

  1. a) Calculate the friction force during braking.
  2. b) Apply the work-energy theorem to calculate the distance it travels until coming to a stop.
  3. c) Reason, from the expression obtained, what happens to the braking distance if the initial speed is doubled: does it also double?
Solve
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On an ice rink, a puck of mass $m$ is launched with an initial speed of $v_0 = 10\ \text{m/s}$, and slides in a straight line until coming to a stop due to friction…

On an ice rink, a puck of mass $m$ is launched with an initial speed of $v_0 = 10\ \text{m/s}$, and slides in a straight line until coming to a stop due to friction ($\mu = 0.25$; take $g = 10\ \text{m/s}^2$).

  1. a) Write the expression for the work done by friction and, applying the work-energy theorem, obtain the stopping distance $d$ as a function of $v_0$, $\mu$, and $g$.
  2. b) Calculate this distance numerically.
  3. c) Reason, from the expression obtained, whether the stopping distance depends on the mass of the puck. What practical implication does this result have for highway safety?
Solve

The work done by weight does not depend on the path

Path of a gondola going up from 1200 m to 1750 m and down to 1550 m: the work done by the weight depends only on the net change in height of 350 m between departure and arrival Net Δh = 350 m base · 1200 m summit · 1750 m middle · 1550 m ascent: +550 m descent: −200 m
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You place a toolbox of $10\ \text{kg}$ onto a shelf located $2.5\ \text{m}$ above the floor (take $g = 9.8\ \text{m/s}^2$). a) Calculate the weight of the box. b)…

You place a toolbox of $10\ \text{kg}$ onto a shelf located $2.5\ \text{m}$ above the floor (take $g = 9.8\ \text{m/s}^2$).

  1. a) Calculate the weight of the box.
  2. b) Calculate the work done by the weight while raising it. Remember that weight points downwards and displacement upwards.
  3. c) What is the minimum work that you do against gravity?
Solve
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A construction crane lowers a beam of $500\ \text{kg}$ from a height of $8\ \text{m}$ to the ground at constant speed (take $g = 9.8\ \text{m/s}^2$). a) Calculate the…

A construction crane lowers a beam of $500\ \text{kg}$ from a height of $8\ \text{m}$ to the ground at constant speed (take $g = 9.8\ \text{m/s}^2$).

  1. a) Calculate the work done by the weight during the descent.
  2. b) Calculate the work done by the tension in the cable.
  3. c) What is the net work? Relate this to the fact that the speed is constant, using the work-energy theorem.
Solve
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A firefighter of $80\ \text{kg}$ with their equipment climbs a fire escape from the ground to a terrace located $18\ \text{m}$ higher up (take $g = 10\ \text{m/s}^2$).…

A firefighter of $80\ \text{kg}$ with their equipment climbs a fire escape from the ground to a terrace located $18\ \text{m}$ higher up (take $g = 10\ \text{m/s}^2$).

  1. a) Calculate the work done by the weight during the ascent.
  2. b) Calculate the work done by the firefighter against gravity.
  3. c) If instead of the fire escape they had climbed a longer ramp to the same terrace, how would that work change? Justify your answer.
Solve
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A cable car cabin with its passengers has a total mass of $800\ \text{kg}$. On the outbound trip, it ascends from the base station, at an altitude of $1200\ \text{m}$,…

A cable car cabin with its passengers has a total mass of $800\ \text{kg}$. On the outbound trip, it ascends from the base station, at an altitude of $1200\ \text{m}$, to the top station, at $1750\ \text{m}$. It then partially descends to an intermediate station located at $1550\ \text{m}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.

base · 1200 m top · 1750 m intermediate · 1550 m W total = ?
  1. a) Calculate the work done by weight during the ascent.
  2. b) Calculate the work done by weight during the partial descent.
  3. c) Calculate the total work done by weight over the entire trip and verify that it depends only on the net difference in altitude between the starting and ending points. What property of the gravitational force does this result highlight?
Solve

Back to the start: for the musher's sled, only the horizontal component of the rope does useful work ($F \cos\alpha$); for the pallet pushed in a straight line, the whole force is parallel to the displacement (maximum work with $\alpha=0^\circ$); and for the gondola, the work done by the weight depends only on the net change in height between departure and arrival, regardless of the intermediate ascents and descents.

Summary

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