Work, Energy and Power – Conservation of Linear Momentum and Collision (Senior 4 Physics) | Ganzaa

Conservation of Linear Momentum and Collision

Lessons

Senior 4 · Physics

Conservation of Linear Momentum and Collision

4 lessons

Lesson 1 of 4

Work, Energy and Power

25 min

Carrying water up the hill

A 20-litre jerrycan of water has a mass of about 20 kg. Carry it up a hill 50 m high and you have done about 9800 J of work against gravity, whether you walk up in ten minutes or run up in three; but running needs more power. Pump the same water up with a hand pump and some of your effort is lost to friction, so the pump's efficiency matters. In S1 and S2 you met work, energy and power for simple cases. This unit builds the full energy and momentum toolkit of mechanics, starting with a careful look at work.

Work done by a constant force

A force does work when its point of application moves in the direction of the force, or has a component in that direction. For a constant force $F$ whose direction makes an angle $\theta$ with the displacement $s$ (Figure 1.1), only the component $F\cos\theta$ along the displacement does work:

W = F s \cos\theta = \vec{F}\cdot\vec{s}

Figure 1.1: Only the component of the force along the displacement does work.

Work is a scalar, measured in joules (1 J = 1 N m when the force and displacement are parallel). Three cases matter:

$0 \le \theta \lt 90^\circ$ : positive work; the force gives energy to the body (the pull on the cart).
$\theta = 90^\circ$ : zero work; for example the upward force of your arm on a jerrycan you carry along level ground, or the tension in the string of a stone whirled in a circle.
$90^\circ \lt \theta \le 180^\circ$ : negative work; the force takes energy from the body (friction, air resistance, brakes).

Definition

The work $W$ done by a constant force is the product of the force and the displacement of its point of application in the direction of the force: $W = Fs\cos\theta$ . Its SI unit is the joule (J). Energy is the capacity to do work, also measured in joules. Example: lifting a 2.0 kg bag through 1.5 m at steady speed needs $W = 2.0 \times 9.8 \times 1.5 = 29\,\mathrm{J}$ .

Worked example 1: pulling a cart

In Figure 1.1, a 150 N pull at 25° above the horizontal moves a cart 40 m along level ground against a friction force of 110 N. Find the work done by the pull, by friction, by the weight and by the normal force, and the net work.

W_{\text{pull}} = 150 \times 40 \times \cos 25^\circ = 5.4 \times 10^{3}\,\mathrm{J}, \qquad W_{\text{friction}} = 110 \times 40 \times \cos 180^\circ = -4.4 \times 10^{3}\,\mathrm{J}

The weight and the normal force are vertical, at 90° to the displacement, so each does zero work.

W_{\text{net}} = 5438 - 4400 = 1.0 \times 10^{3}\,\mathrm{J}

The net work of about 1000 J goes into the cart's kinetic energy (lesson 2).

Work done by a variable force

If the force changes as the body moves, divide the motion into small steps $\Delta s$ , in each of which $F$ is almost constant: the work is the sum of the small rectangles $F\Delta s$ , which is the area under the force–displacement graph. For a spring obeying Hooke's law, $F = k x$ (Figure 1.2), the area is a triangle:

W = \tfrac{1}{2}F x = \tfrac{1}{2}k x^2

Figure 1.2: The work done by a variable force is the area under the force–displacement graph.

This work is stored in the stretched spring as elastic potential energy, the same result you met for a stretched wire in Unit 3.

Worked example 2: stretching a spring

A spring of stiffness $800\,\mathrm{N\,m^{-1}}$ is stretched by 0.15 m. Find the work done, and the extra work needed to stretch it a further 0.05 m.

W_1 = \tfrac{1}{2} \times 800 \times 0.15^2 = 9.0\,\mathrm{J}, \qquad W_2 = \tfrac{1}{2} \times 800 \times 0.20^2 - 9.0 = 16 - 9.0 = 7.0\,\mathrm{J}

The last 5 cm needs almost as much work as the first 15 cm, because the force is larger.

Power and efficiency

Power is the rate of doing work (or of transferring energy):

P = \frac{W}{t}, \qquad \text{unit: watt (W)} = 1\,\mathrm{J\,s^{-1}}

For a force $F$ moving its point of application at velocity $v$ in its own direction, in a short time $\Delta t$ it does work $F\Delta s$ , so

P = \frac{F\Delta s}{\Delta t} = F v

This is the power a vehicle's engine must deliver to keep moving at speed $v$ against a total opposing force $F$ . No machine converts all its input energy into useful output; the rest is wasted, mostly as internal (thermal) energy. The efficiency is

\eta = \frac{\text{useful energy output}}{\text{total energy input}} \times 100\% = \frac{\text{useful power output}}{\text{total power input}} \times 100\%

Electrical energy is often billed in kilowatt-hours: $1\,\mathrm{kWh} = 1000\,\mathrm{W} \times 3600\,\mathrm{s} = 3.6 \times 10^{6}\,\mathrm{J}$ .

Definition

Power is the rate at which work is done: $P = \frac{W}{t}$ , and for a force moving at velocity $v$ , $P = F v$ . Its unit is the watt (W), one joule per second. Efficiency is the ratio of useful output energy (or power) to total input energy (or power). Example: a 60 W bulb transfers 60 J of energy every second.

Worked example 3: a car on a hill

A car of mass 1200 kg climbs a road that rises 1 m for every 20 m travelled along it, at a steady 15 m s⁻¹. Air resistance and friction total 500 N. Find the power the engine delivers to the wheels.

At steady speed the driving force balances the component of weight down the slope plus the resistance. With $\sin\theta = \frac{1}{20} = 0.050$ :

F = mg\sin\theta + 500 = 1200 \times 9.8 \times 0.050 + 500 = 588 + 500 = 1088\,\mathrm{N}

P = Fv = 1088 \times 15 = 1.6 \times 10^{4}\,\mathrm{W} = 16\,\mathrm{kW}

Worked example 4: the efficiency of a pump

An electric pump raises 1500 kg of water through a height of 12 m into a tank in 5.0 minutes. Its electrical input power is 800 W. Find its efficiency.

P_{\text{useful}} = \frac{mgh}{t} = \frac{1500 \times 9.8 \times 12}{300} = 588\,\mathrm{W}, \qquad \eta = \frac{588}{800} \times 100\% = 74\%

The other 212 W heats the motor and the water and is lost to friction in the pipes.

Experiment 1.1: your own power

Apparatus: a flight of stairs, a metre rule, a stopwatch, a bathroom scale.

Measure your mass $m$ . Measure the height of one step and count the steps to find the total vertical height $h$ .
A partner times how long, $t$ , you take to walk up the stairs, and then to run up them safely.
Calculate the work done against gravity, $m g h$ , and your power $P = \frac{mgh}{t}$ for walking and for running.

A 60 kg student who runs up a 3.0 m flight in 4.0 s develops $\frac{60 \times 9.8 \times 3.0}{4.0} = 441\,\mathrm{W}$ , roughly half a horsepower; walking up takes longer and needs less power, for the same work. The calculation ignores the energy used to move the legs and to speed up, so the body's actual power output is larger.

Warning

Safety. Run only on dry stairs with a handrail, one student at a time, and stop if you feel dizzy. Nobody should be pushed to compete.

Warning

Common mistakes. (1) Using $W = F s$ when the force is at an angle to the displacement: use $Fs\cos\theta$ . (2) Saying a person holding a heavy load still is doing work on it: the load does not move, so no work is done on it. (3) Confusing energy (J) with power (W). (4) Forgetting to convert minutes to seconds and kWh to joules.

Tip

Every work question: draw the force and the displacement, find the angle between them, then $W = Fs\cos\theta$ . If the force changes, find the area under the $F$ – $s$ graph.

Work and power (Bozeman Science). Paul Andersen explains work as force times displacement in the direction of the force, and power as the rate of doing work.

✦

Practice

A box is pushed 8.0 m across a floor by a horizontal force of 45 N. How much work is done by the push?
A child pulls a toy car 12 m with a string at 40° to the horizontal. The tension is 6.0 N. Find the work done by the tension.
A 70 kg climber ascends 450 m. How much work does she do against gravity? If she takes 2.5 h, what is her average power output against gravity?
A spring of stiffness $250\,\mathrm{N\,m^{-1}}$ is compressed by 4.0 cm. Find the energy stored.
A force on a body rises steadily from 0 to 30 N over the first 2.0 m and stays at 30 N for the next 3.0 m. Find the total work done.
A motorcycle and rider of total mass 220 kg travel at a steady 20 m s⁻¹ on a level road against a resistance of 180 N. Find the power output. What power is needed to climb a slope with $\sin\theta = 0.10$ at the same speed (same resistance)?
A 2.0 kW electric kettle takes 4.0 min to boil water. How many kWh does it use, and what does that cost at 250 Frw per kWh (illustrative tariff)?
A crane motor of input power 15 kW lifts a 1.2 t load at 0.80 m s⁻¹. A technician says the crane is 80% efficient. Evaluate the claim.

Show answersHide answers

$W = 45 \times 8.0 = 360\,\mathrm{J}$ .
$W = 6.0 \times 12 \times \cos 40^\circ = 55\,\mathrm{J}$ .
$W = 70 \times 9.8 \times 450 = 3.1 \times 10^{5}\,\mathrm{J}$ ; $P = \frac{3.09 \times 10^{5}}{9000} = 34\,\mathrm{W}$ .
$E = \tfrac{1}{2} \times 250 \times 0.040^2 = 0.20\,\mathrm{J}$ .
Area: triangle $\tfrac{1}{2} \times 2.0 \times 30 = 30\,\mathrm{J}$ plus rectangle $3.0 \times 30 = 90\,\mathrm{J}$ ; total 120 J.
Level: $P = 180 \times 20 = 3.6\,\mathrm{kW}$ . Slope: $F = 220 \times 9.8 \times 0.10 + 180 = 396\,\mathrm{N}$ , $P = 396 \times 20 = 7.9\,\mathrm{kW}$ .
$2.0\,\mathrm{kW} \times \frac{4.0}{60}\,\mathrm{h} = 0.13\,\mathrm{kWh}$ ; cost about $0.133 \times 250 = 33\,\mathrm{Frw}$ .
Useful power $= mgv = 1200 \times 9.8 \times 0.80 = 9.4\,\mathrm{kW}$ ; $\eta = \frac{9.4}{15} = 63\%$ . The claim of 80% is too high (unless the input power figure is wrong).

Quick check

Key points

$W = Fs\cos\theta$ : positive, zero or negative depending on the angle between force and displacement.
For a variable force, work = area under the $F$ – $s$ graph; for a spring $W = \tfrac{1}{2}kx^2$ .
$P = \frac{W}{t} = Fv$ ; efficiency $= \frac{\text{useful output}}{\text{input}} \times 100\%$ ; $1\,\mathrm{kWh} = 3.6 \times 10^{6}\,\mathrm{J}$ .

Sources and credits

7.1: Work: The Scientific Definition and 7.7: Power, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Children at the hand pump.jpg by Emerymakini, Wikimedia Commons, licensed under CC BY 4.0.
Photo: Stefanie Dröxler clean and jerk-4316.jpg by Isiwal, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Work and Power, Bozeman Science, embedded from YouTube.

The diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1Carrying water up the hill

2Work done by a constant force

3Work done by a variable force

4Power and efficiency

5Experiment 1.1: your own power

Carrying water up the hill

Work done by a constant force

Work done by a variable force

Power and efficiency

Experiment 1.1: your own power