Work, energy and power

Physics · Senior school physics · about 10 min

Work tells us how a force transfers energy over a distance, energy helps us track changes in motion and height, and power tells us how fast energy is transferred. These ideas often solve motion questions more directly than resolving every force over time.

The key idea

The net work done on an object equals its change in kinetic energy:

Wnet=ΔK=12m(vf2−vi2)W_{\mathrm{net}}=\Delta K=\frac12 m\left(v_f^2-v_i^2\right)

Here mm is mass, viv_i and vfv_f are initial and final speeds, and WnetW_{\mathrm{net}} is the sum of the work done by all forces.

Draw a straight ramp descending from upper left to lower right, with a small wheeled cart near its top and an arrow pointing downhill along the ramp to indicate travel. Draw a short resistive-force arrow parallel to the ramp pointing uphill from the cart, and a vertical gravity arrow mg pointing down from its centre. Place the start at the ramp top and finish at the bottom. To the right of the ramp, draw a vertical dimension line from the start elevation down to the finish elevation labelled h; along the ramp draw a parallel dimension line labelled d. Near the finish, add a tangential velocity arrow pointing downhill labelled v. No numerical results.
Gravity supplies energy through vertical drop, while resistance does negative work along the path.

Work depends on direction

For a constant force FF and straight displacement dd, the work done is W=Fdcos⁡θW=Fd\cos\theta, where θ\theta is the angle between force and displacement. A force along the motion does positive work; a force opposite it does negative work. A perpendicular force does no work. Work is measured in joules: 1 J=1 N m1\,\mathrm{J}=1\,\mathrm{N\,m}. The work-energy theorem uses net work, so include gravity, resistance and any applied forces as appropriate. Avoid multiplying a force by the total path length if that force does not act along the path.

Track kinetic and potential energy

Kinetic energy is K=12mv2K=\tfrac12 mv^2. Near Earth's surface, gravitational potential energy relative to a chosen reference height is Ug=mghU_g=mgh. If only gravity does work, Ki+Ui=Kf+UfK_i+U_i=K_f+U_f. With resistance, mechanical energy decreases by the positive magnitude of the energy dissipated. Equivalently, Ki+Ui+Wnc=Kf+UfK_i+U_i+W_{\mathrm{nc}}=K_f+U_f, where WncW_{\mathrm{nc}} is the work of non-conservative forces and is negative for friction. Energy is conserved overall, but some mechanical energy becomes thermal energy. State the reference height consistently.

Power measures the rate

Average power is energy transferred divided by elapsed time, Pavg=ΔE/ΔtP_{\mathrm{avg}}=\Delta E/\Delta t. Its SI unit is the watt, with 1 W=1 J/s1\,\mathrm{W}=1\,\mathrm{J/s}. For a moving object, the instantaneous power delivered by a force is P=F⋅vP=\mathbf F\cdot\mathbf v. The force must be taken along the velocity direction for positive delivered power. A resistive force does negative mechanical work, but the rate of energy dissipated is usually reported as a positive number. Check what the question means by power before assigning a sign.

Worked example: A cart descending a resistive ramp

A 200 kg200\,\mathrm{kg} cart starts from rest and travels down a straight 10.0 m10.0\,\mathrm{m} ramp with a vertical drop of 5.0 m5.0\,\mathrm{m}. A constant 100 N100\,\mathrm{N} resistive force acts up the slope. The descent takes 2.13 s2.13\,\mathrm{s}. Using g=9.81 m/s2g=9.81\,\mathrm{m/s^2}, find the speed at the bottom and the average rate at which resistance dissipates mechanical energy.

Show a straight right-descending incline that is 10.0 m long along its surface and drops 5.0 m vertically from left top to right bottom, so the ramp angle is 30 degrees to horizontal. A wheeled cart near the top has its mass written inside it. Draw an arrow for motion down the slope and an opposing resistance arrow along the slope pointing back towards the top, labelled by its force magnitude. Use dimension arrows along the inclined surface for 10.0 m and vertically from starting elevation to bottom elevation for 5.0 m. Label the high and low ends, with no final speed or power numerical value.
The drop is vertical, but resistance acts over the full sloping distance.
Show the solution

Choose the bottom of the ramp as zero gravitational potential energy. Resistance acts opposite the motion.

  1. Gravity supplies potential energy equal to
ΔUg=mg(hf−hi)=200(9.81)(0−5.0)=−9810 J.\Delta U_g=mg(h_f-h_i)=200(9.81)(0-5.0)=-9810\,\mathrm{J}.

Thus gravity does Wg=9810 JW_g=9810\,\mathrm{J} of positive work.

  1. Resistance does negative work along the ramp:
Wr=−Fd=−100(10.0)=−1000 J.W_r=-Fd=-100(10.0)=-1000\,\mathrm{J}.
  1. The cart starts from rest, so the net work becomes final kinetic energy:
12mvf2=Wg+Wr=9810−1000=8810 J.\frac12 mv_f^2=W_g+W_r=9810-1000=8810\,\mathrm{J}. vf=2(8810)200=88.1=9.386… m/s.v_f=\sqrt{\frac{2(8810)}{200}}=\sqrt{88.1}=9.386\ldots\,\mathrm{m/s}.
  1. The positive rate of energy dissipation by resistance is
Ploss,avg=∣Wr∣Δt=10002.13=469.48… W.P_{\mathrm{loss,avg}}=\frac{|W_r|}{\Delta t}=\frac{1000}{2.13}=469.48\ldots\,\mathrm{W}.

Answer: bottom speed 9.39 m/s9.39\,\mathrm{m/s} and average energy dissipation rate 469 W469\,\mathrm{W}.

Common mistakes

Assuming work is always positive

Use the angle between force and displacement. Opposing resistance does negative work.

Conserving mechanical energy despite friction

Include the non-conservative work term or subtract dissipated energy.

Using path length instead of vertical drop in mghmgh

Use the change in vertical height for gravitational potential energy.

Confusing energy with power

Energy is in J\mathrm{J}; divide by time in seconds to obtain average power in W\mathrm{W}.

Check your understanding

1. A 30 N30\,\mathrm{N} constant force acts at 60∘60^\circ to a 4.0 m4.0\,\mathrm{m} straight displacement. How much work does it do?

2. A 2.0 kg2.0\,\mathrm{kg} object speeds up from 3.0 m/s3.0\,\mathrm{m/s} to 5.0 m/s5.0\,\mathrm{m/s}. What is the net work done?

3. A machine transfers 900 J900\,\mathrm{J} in 3.0 s3.0\,\mathrm{s}. What is its average power?

Still not sure? Talk it through.

Waypoint Mentor's tutor explains this out loud, works through your own problems on a whiteboard, and checks you've got it. Free to start.

Explain this to me

Related topics