Q.Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m ($g = 10\ \mathrm{m\,s^{-2}}$).

Step 1. The force is applied vertically upward to lift the load, and the displacement (10 m) is also vertically upward, so the angle between them is $\theta=0^\circ$, $\cos\theta=1$.

Step 2. Given: $F=30$ N, $d=10$ m.

Step 3. Work done by this force: $W = Fd\cos\theta = (30)(10)(1) = 300\ \mathrm{J}$.

Step 4. Note the 2 kg mass and $g=10\ \mathrm{m\,s^{-2}}$ are not needed to compute the work done by the stated 30 N force -- that quantity depends only on the force actually applied and the distance it acts through; the mass/weight would matter only if we were asked for the net work done on the load (which would separately need to subtract the work done against gravity, $mgh=20\times...$ wait $=1\times2\times10\times10$, i.e. $mgh=(2)(10)(10)=200$ J).

Multiply the given applied force by the vertical distance it acts through (angle zero, so cos(theta) = 1).

• Trying to substitute the load's weight (mg = 20 N) for the given applied force (30 N) -- the question asks for work done BY THE STATED 30 N FORCE.
• Unnecessarily subtracting or adding the gravitational work when only the work done by the applied force is asked for.