Question

6. A 5 kg bowling ball is dropped from the top of a 50 m high building.
a.How much potential energy does the bowling ball have before it is dropped?
b.How fast will it be moving when it hits the ground?

Answer 1

Explanations & Calculations

• At a given instance the mechanical energy of an object is given by the sum of potential( $E_p$ ) & kinetic ($E_k$) energies.
• This mechanical energy is conserved at the absence of external forces that may act on that object.

a) Potential energy it had at the top with respect to the ground level

$\qquad\qquad \begin{aligned} \small E_p &= \small mgh\\ &= \small 5kg\times 9.8ms^{-2}\times 50m\\ &=\small \bold{2450J} \end{aligned}$

b) Mechanical energy is conserved through the top of the building to the ground if the air resistance is neglected. Therefore,

$\qquad\qquad \begin{aligned} \small (E_p+E_k)_{top} &= \small (E_p+E_k)_{ground}\\ \small 2450J+0&= \small 0J+\frac{1}{2}mv^2\\ \small 2450&= \small \frac{1}{2}\times5kg\times v^2\\ \small v &= \small \bold{31.30ms^{-1}} \end{aligned}$