Work, Energy and Power: Question 5

Worked solution

Part (a): Loss in gravitational potential energy

The loss in gravitational PE is given by $\Delta E_p = mg\Delta h$: $\Delta E_p = 1.2\times9.81\times1.8$

Working in two steps: $1.2\times9.81=11.772$, then $11.772\times1.8=21.1896\text{ J}$.

Check by recomputing the second step differently: $11.772\times1.8 = 11.772\times2 - 11.772\times0.2 = 23.544-2.3544=21.1896\text{ J}$. Both routes agree.

So the loss in gravitational PE is $\boxed{21.2}\text{ J}$ (3 s.f.).

Part (b): Kinetic energy just before impact

Since air resistance is negligible, no energy is lost to other forms (such as heat) as the book falls. The principle of conservation of energy means that all of the gravitational PE lost is converted directly into kinetic energy. Therefore: $E_k = \Delta E_p = \boxed{21.2}\text{ J}$

Part (c): Speed just before impact

Using $E_k=\tfrac12mv^2$, rearranged for $v$: $v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2\times21.1896}{1.2}}$

Working the numerator first: $2\times21.1896=42.3792$. Dividing by the mass: $42.3792/1.2=35.316$.

$v = \sqrt{35.316} \approx 5.94\text{ m s}^{-1}$

Check using an independent route: for free fall from rest with no air resistance, $v=\sqrt{2g\Delta h}=\sqrt{2\times9.81\times1.8}=\sqrt{35.316}\approx5.94\text{ m s}^{-1}$, the same result, confirming the answer.

Final answers

• (a) Loss in gravitational PE $= \boxed{21.2}\text{ J}$
• (b) Kinetic energy just before impact $= \boxed{21.2}\text{ J}$ (equal to the PE lost, by conservation of energy)
• (c) Speed just before impact $\approx \boxed{5.94}\text{ m s}^{-1}$