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Prep 1 · Physics and Chemistry
Energy · Work and mechanical energy

Work and Power

Before you start, make sure you know: Work and power (Secondary Year 2)
Before you start, make sure you know: Mechanical energy: kinetic and potential (Secondary Year 2)

Ximena pulls her rolling suitcase along the Metro platform. If she pulls the handle horizontally, it moves along easily; if she pulls at a $45^\circ$ angle, she feels that part of her effort goes upward instead of forward. With the same force, how much energy does she actually transfer to the suitcase by tilting the handle? You'll calculate it at the end of the page.

Work transfers energy

In physics, a force does work when, acting on a body that moves, it gives energy to the body or takes energy from it. It is a way of transferring energy from one object to another: your hand gives energy to the box you push, and the friction of the floor takes it away.

A person pushes a box with a force F over a distance d; the work W transfers energy to the box. F distance d W = energy transferred

Work done by a constant force: $W = F \cdot d \cdot \cos\alpha$

When the force and the displacement point in the same direction, you just multiply: $W = F \cdot d$. If the force is tilted, only the part of the force that points in the direction of motion transfers energy.

If you're given the work and asked for the force, solve for it: with the force parallel to the motion, $F = \dfrac{W}{d}$. With an angle, $F = \dfrac{W}{d\cos\alpha}$.

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A worker pushes a pallet with a horizontal force of $150\ \text{N}$ and moves it $8\ \text{m}$ along the warehouse floor. Calculate the work done.

A worker pushes a pallet with a horizontal force of $150\ \text{N}$ and moves it $8\ \text{m}$ along the warehouse floor. Calculate the work done.

Solve
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A child pulls a cart with a rope that forms an angle of $60^\circ$ with the horizontal, exerting a force of $200\ \text{N}$, and drags it $10\ \text{m}$ (…

A child pulls a cart with a rope that forms an angle of $60^\circ$ with the horizontal, exerting a force of $200\ \text{N}$, and drags it $10\ \text{m}$ ($\cos 60^\circ = 0.5$), as shown in the figure.

F = 200 N 60° d = 10 m W = ?

Calculate the work done.

Solve
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To drag a piano $3\ \text{m}$ along the stage floor, $900\ \text{J}$ of work was done. With what average horizontal force was it pushed? Rearrange the formula for work…

To drag a piano $3\ \text{m}$ along the stage floor, $900\ \text{J}$ of work was done. With what average horizontal force was it pushed? Rearrange the formula for work to solve for force first.

Solve

Three important cases: 0°, 90°, and 180°

Three cases of the work done by a force on a block depending on the angle with the displacement: 0° (maximum positive work), 90° (zero work), and 180° (negative work). α = 0°, cos α = 1 α = 90°, cos α = 0 α = 180°, cos α = −1 F F F displacement displacement displacement maximum W > 0 W = 0 W < 0
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The work done by a force on a moving object is calculated as $W = F \times d \times \cos \alpha$, where $\alpha$ is the angle between the force and the displacement. You…

The work done by a force on a moving object is calculated as $W = F \times d \times \cos \alpha$, where $\alpha$ is the angle between the force and the displacement. You pull a trolley with the same force and over the same distance in three cases: with the rope horizontal ($\alpha = 0^\circ$), at an angle of $45^\circ$, and at an angle of $90^\circ$ ($\cos 0^\circ = 1$; $\cos 45^\circ \approx 0.71$; $\cos 90^\circ = 0$). The figure shows the three cases.

F α = 0° F α = 45° F α = 90°

Order the three amounts of work done from greatest to least and justify your answer using the formula.

Solve
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A satellite orbits the Earth in a circular orbit at a constant speed, as shown in the figure. The gravitational force exerted on it by the Earth points toward the center…

A satellite orbits the Earth in a circular orbit at a constant speed, as shown in the figure. The gravitational force exerted on it by the Earth points toward the center of the orbit, while its displacement is tangent to the trajectory.

Earth satellite F v
  1. a) What angle do the force and the displacement form?
  2. b) How much work does gravity do on the satellite during each orbit? Justify your answer using the cosine.
Solve
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Aitana mows the lawn by pushing the lawnmower with a force of $80\ \text{N}$ applied to the handle, which forms an angle of $30^\circ$ with the horizontal (…

Aitana mows the lawn by pushing the lawnmower with a force of $80\ \text{N}$ applied to the handle, which forms an angle of $30^\circ$ with the horizontal ($\cos 30^\circ \approx 0.87$), and covers $20\ \text{m}$ of lawn, as shown in the figure.

F = 80 N 30° Fₓ d = 20 m W = ?
  1. a) Calculate the work done.
  2. b) What work would have been done by pushing horizontally with the same force? What accounts for the difference?
Solve

Work against weight: $W = m\,g\,h$

Lifting a box onto a shelf is the most common kind of work. To lift it at constant speed, your force has to equal its weight, $F = m\,g$, and the vertical displacement is the height $h$. Force and displacement point in the same direction ($\alpha = 0^\circ$), so:

A box goes up from the floor to a shelf at height h: the upward force equals the weight m·g. F = m g weight height h
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You place a box of books of $8\ \text{kg}$ on the top shelf of a cupboard, $1.5\ \text{m}$ above the floor (take $g = 10\ \text{m/s}^2$). a) What minimum force do you…

You place a box of books of $8\ \text{kg}$ on the top shelf of a cupboard, $1.5\ \text{m}$ above the floor (take $g = 10\ \text{m/s}^2$).

  1. a) What minimum force do you have to overcome to raise it?
  2. b) What work do you do against gravity?
Solve
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A crane lifts a steel beam of $25\ \text{kg}$ to a height of $12\ \text{m}$ at a constant speed (take $g = 9.8\ \text{m/s}^2$). a) What force does the cable exert? b)…

A crane lifts a steel beam of $25\ \text{kg}$ to a height of $12\ \text{m}$ at a constant speed (take $g = 9.8\ \text{m/s}^2$).

  1. a) What force does the cable exert?
  2. b) What work does the crane do?
Solve
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In a weightlifting competition, an athlete lifts a weight of $100\ \text{kg}$ from the floor to a height of $2\ \text{m}$ in $2\ \text{s}$, holds it still above her head…

In a weightlifting competition, an athlete lifts a weight of $100\ \text{kg}$ from the floor to a height of $2\ \text{m}$ in $2\ \text{s}$, holds it still above her head for $3\ \text{s}$, and then walks $10\ \text{m}$ horizontally with it until leaving the platform (take $g = 10\ \text{m/s}^2$). The figure shows the three phases.

1 · lifts 2 · holds still 3 · walks F 2 m F still 3 s F 10 m W = ? W = ? W = ?
  1. a) What work does she do against gravity while lifting it?
  2. b) What power does she develop during this lift?
  3. c) Does she do work against gravity while holding it still? What about while walking horizontally? Justify both answers using the definition of work.
Solve

Friction and net work

Several forces usually act on a body. The net work is the sum of the work done by all of them (with their signs). It equals the work done by the net force.

Friction always opposes the motion ($\alpha = 180^\circ$), so its work is negative: $W_{fric} = -f \cdot d$. That energy doesn't disappear: it turns into heat that warms the surfaces.

When there are several stretches, calculate the work for each one and add: $W_{total} = W_1 + W_2 + \dots$

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A worker pushes a container $12\ \text{m}$ along a warehouse floor with a horizontal force of $80\ \text{N}$, while friction opposes it with $45\ \text{N}$, as shown in…

A worker pushes a container $12\ \text{m}$ along a warehouse floor with a horizontal force of $80\ \text{N}$, while friction opposes it with $45\ \text{N}$, as shown in the figure. Remember that work done by a force opposing displacement is negative ($\alpha = 180^\circ$, $\cos 180^\circ = -1$).

container F = 80 N friction = 45 N d = 12 m W = ?
  1. a) Calculate the work done by the worker.
  2. b) Calculate the work done by friction.
  3. c) What is the net work done on the container?
Solve
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A kayaker drags his kayak $40\ \text{m}$ across the sand by pulling a rope that forms an angle of $45^\circ$ with the horizontal, with a force of $50\ \text{N}$ (…

A kayaker drags his kayak $40\ \text{m}$ across the sand by pulling a rope that forms an angle of $45^\circ$ with the horizontal, with a force of $50\ \text{N}$ ($\cos 45^\circ \approx 0.71$). Friction with the sand opposes the motion with $20\ \text{N}$.

  1. a) Calculate the work done by the kayaker.
  2. b) Calculate the work done by friction (remember it is negative).
  3. c) What is the net work done on the kayak? How can you observe the effect of this net work?
Solve
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A delivery worker drags a parcel trolley in two stages. In the first stage, he pulls the rope, which forms an angle of $30^\circ$ with the horizontal, with a force of…

A delivery worker drags a parcel trolley in two stages. In the first stage, he pulls the rope, which forms an angle of $30^\circ$ with the horizontal, with a force of $40\ \text{N}$ for $20\ \text{m}$ ($\cos 30^\circ \approx 0.87$); in the second stage, the rope is horizontal and he pulls with $40\ \text{N}$ for $15\ \text{m}$.

  1. a) Calculate the work done in each stage and the total work done.
  2. b) If he takes $36\ \text{s}$ for the entire distance, what average power does he develop?
Solve

Power: how fast energy is transferred

Two forklifts lift a pallet of bricks with a mass of $400\ \text{kg}$ to a height of $10\ \text{m}$. Both do the same work, $m\,g\,h = 400 \times 10 \times 10 = 40{,}000\ \text{J}$ (with $g = 10\ \text{m/s}^2$), but A takes $8\ \text{s}$ and B takes $40\ \text{s}$. How do we measure which one transfers energy faster?

For forklift A: $P = 40{,}000/8 = 5{,}000\ \text{W}$. For B: $40{,}000/40 = 1{,}000\ \text{W}$. Same work, but A is five times more powerful because it does it in less time.

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The motor of a conveyor belt in a parcel distribution center does $18000\ \text{J}$ of work in $60\ \text{s}$. Power is calculated as $P = W/t$. Calculate the power of…

The motor of a conveyor belt in a parcel distribution center does $18000\ \text{J}$ of work in $60\ \text{s}$. Power is calculated as $P = W/t$. Calculate the power of the motor, in watts.

Solve
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Classify the following units according to whether they measure work (or energy) or power, and explain the criterion you used: a) $\text{J}$ b) $\text{W}$ c)…

Classify the following units according to whether they measure work (or energy) or power, and explain the criterion you used:

  1. a) $\text{J}$
  2. b) $\text{W}$
  3. c) $\text{N} \times \text{m}$
  4. d) $\text{kWh}$
  5. e) $\text{kW}$
Solve
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A mountaineer and their backpack have a combined mass of $75\ \text{kg}$ and gain $720\ \text{m}$ of elevation in $1.5\ \text{h}$ (take $g = 10\ \text{m/s}^2$). a) What…

A mountaineer and their backpack have a combined mass of $75\ \text{kg}$ and gain $720\ \text{m}$ of elevation in $1.5\ \text{h}$ (take $g = 10\ \text{m/s}^2$).

  1. a) What force is overcome when climbing?
  2. b) What work is done against weight?
  3. c) What average power is developed? Remember to convert time into seconds.
Solve

From watts to the kilowatt-hour

If you solve the power formula for work, you get $W = P \cdot t$: the energy an appliance uses or transfers is its power times the time it runs.

According to what they measure, the units are classified as follows: $\text{J}$, $\text{N}\cdot\text{m}$, and $\text{kWh}$ measure energy or work; $\text{W}$, $\text{kW}$, and $\text{hp}$ measure power.

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A drill has a power rating of $600\ \text{W}$ and operates for $2\ \text{min}$ ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$). a) Express the time in seconds. b)…

A drill has a power rating of $600\ \text{W}$ and operates for $2\ \text{min}$ ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).

  1. a) Express the time in seconds.
  2. b) Calculate the work (energy) done.
  3. c) Express this energy in $\text{kWh}$.
Solve
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A swimming pool filter pump has a power rating of $750\ \text{W}$ and operates for $8\ \text{h}$ a day ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$). a) Calculate the…

A swimming pool filter pump has a power rating of $750\ \text{W}$ and operates for $8\ \text{h}$ a day ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).

  1. a) Calculate the energy consumed per day, in joules.
  2. b) Express it in $\text{kWh}$, the unit used by the electricity supplier for billing.
Solve
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The human heart does about $1.2\ \text{J}$ of work with each beat ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$). a) Calculate its average power at rest, with $60$ beats…

The human heart does about $1.2\ \text{J}$ of work with each beat ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).

  1. a) Calculate its average power at rest, with $60$ beats per minute.
  2. b) How much energy does it pump in a full day? Express it in joules and in $\text{kWh}$.
  3. c) Compare it with the power of a $10\ \text{W}$ LED light bulb and comment on the result.
Solve

Useful power, efficiency, and the advantage of the ramp

The power a motor consumes (for example, what it draws from the electrical grid) is always greater than the useful power it delivers to the load. In a real machine, part of the energy is lost as heat and sound due to friction between parts and the resistance of the wires.

A motor receives a total power; part of it becomes useful power that lifts the load and the rest is lost as heat and sound. motor total P useful P: lifts the load losses: heat and sound

That is why a $150\ \text{kW}$ ($\approx 201\ \text{hp}$) motor in a cargo truck climbing a hill delivers much less useful power on the climb than the power it consumes: part is lost as heat in the engine and another part to friction with the air and the tires. The maximum power it can reach is a separate matter: that margin is not energy lost.

The ramp: same energy, less force

The ramp doesn't "save" energy: it trades a large force over a short distance for a small force over a long distance. With real friction, you also have to spend a bit more.

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A 4x4 car of $1800\ \text{kg}$ drives up a mountain pass gaining $100\ \text{m}$ in altitude in $2\ \text{min}$, at constant speed (take $g = 10\ \text{m/s}^2$). a)…

A 4x4 car of $1800\ \text{kg}$ drives up a mountain pass gaining $100\ \text{m}$ in altitude in $2\ \text{min}$, at constant speed (take $g = 10\ \text{m/s}^2$).

  1. a) Calculate the work done against gravity.
  2. b) Calculate the useful power used to climb, in $\text{kW}$.
  3. c) The car's engine specifies a power rating of $150\ \text{kW}$: explain at least two reasons why the useful power you calculated is much lower.
Solve
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In an automated warehouse, a robot lifts packages of $5\ \text{kg}$ onto a shelf $6\ \text{m}$ high, at a rate of $120$ packages per hour (take $g = 10\ \text{m/s}^2$).…

In an automated warehouse, a robot lifts packages of $5\ \text{kg}$ onto a shelf $6\ \text{m}$ high, at a rate of $120$ packages per hour (take $g = 10\ \text{m/s}^2$).

  1. a) What work does it do with each package?
  2. b) What total work does it do in one hour?
  3. c) What average power does it develop?
  4. d) The robot's motor is rated at $500\ \text{W}$, considerably more than what you calculated: give reasons why this margin is needed.
Solve
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To load a box of $60\ \text{kg}$ onto a lorry, whose bed is $1.2\ \text{m}$ above the ground, you can lift it vertically or push it up a frictionless ramp $3\ \text{m}$…

To load a box of $60\ \text{kg}$ onto a lorry, whose bed is $1.2\ \text{m}$ above the ground, you can lift it vertically or push it up a frictionless ramp $3\ \text{m}$ long (take $g = 10\ \text{m/s}^2$). The figure shows the two options.

lorry 1.2 m F vertically F 3 m ramp W = ?
  1. a) Calculate the work required in each case and explain why it is the same.
  2. b) Calculate the force that must be applied in each case.
  3. c) What is the advantage of using the ramp? What is the "cost" of using it?
Solve

Summary

Go back to the beginning: Ximena's suitcase. With the handle at $45^\circ$, only $\cos 45^\circ \approx 0.71$ of the force pushes forward: she transfers about $71\,\%$ of the energy she would transfer pulling horizontally.

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