Physic Labs

Newtonian mechanics

Work and power

Constant-force work is the force component along displacement times displacement; power measures the rate of doing work.

Pushing a wall that does not move can be tiring, but your force does no mechanical work on the wall. Work depends on force, displacement, and their angle; power tells how quickly work is done.

W=Fscos⁡θ,P=WΔtW=Fs\cos\theta,\qquad P=\frac{W}{\Delta t}

Definition: Work and power

For constant force F⃗\vec F and displacement s⃗\vec s, work is W=Fscos⁡θW=Fs\cos\theta (joules, J), where θ\theta is their angle. Average power is P=W/ΔtP=W/\Delta t (watts, W); instantaneous power is the rate of work. A force perpendicular to displacement does zero work.

Vary force, displacement, and pulling angle; see signed work and power update live.

The sign of work

A force component along the displacement does positive work and transfers energy to the object. An opposing component does negative work, often removing mechanical energy. For motion with velocity parallel to the force, instantaneous power is P=F⃗⋅v⃗=FvP=\vec F\cdot\vec v=Fv; power is not the same as work.

Example: Pulling a cart

A 20 N horizontal force pulls a cart 3 m in 4 s. Find the work and average power.

Solution

W=Fs=60 JW=Fs=60\,J; P=W/Δt=60/4=15 WP=W/\Delta t=60/4=15\,W.

The unit of work is the joule: a 1 N force moving an object 1 m along its direction does 1 J. Carrying a bag horizontally at constant height gives zero mechanical work by the upward supporting force because it is perpendicular to displacement (even though the body still uses biological energy). Power distinguishes how quickly equal work is done: 600 J completed in 3 s corresponds to 200 W.

Example: Climbing stairs

A 40 kg student climbs 2 m in 4 s. Take g=10 m/s². Find work against gravity and average power.

Solution

W=mgh=40×10×2=800 J; P=W/t=800/4=200 W.

Work and power use different units: a joule measures transferred energy, while a watt measures the rate of transfer. A higher-powered motor can do the same work faster, but that does not necessarily mean it supplies more energy over a different time interval. Keep time in seconds when using P=W/tP=W/t.

Quick check

How much work does a force perpendicular to displacement do?

For the same work completed in less time, what happens to average power?

References

  1. Young, Freedman (2019). University Physics