Question

Two boys $\text{A}$ and $\text{B}$ are standing at two positions on a staircase. $\text{B}$ has a $500\text{g}$ ball in his hand which is $50\text{cm}$ higher than the platform he is on. Each step is $20\text{cm}$ high and $20\text{cm}$ wide.

$\text{B}$ drops the ball from his hand down the stairs and $\text{A}$ stops the ball on the second step. Find the change in potential energy of the ball.


Consider $\text{g}$ $=10{\text{m s}}^{-2}$ and assume that there is no loss of energy when the ball rolls down the staircase.

• A. $+8.5\text{J}$
• B. $+6.0\text{J}$
• C. $-6.0\text{J}$
• D. $-8.5\text{J}$

Answer 1

The correct option is D $-8.5\text{J}$

Given that:
Height of each step $=20\text{cm}$
Width of each step $=20\text{cm}$
$\text{g}=10{\text{m s}}^{-2}$

Assume potential energy to be zero at the ground.

Initial position of the ball with respect to the ground,
${\text{h}}_i=50+(20\times 8)=210\text{cm}$

Final position of the ball with respect to the ground,
${\text{h}}_f=$ Level of the second step above the ground
${\text{h}}_f=(20\times 2)=40\text{cm}=0.4\text{m}$

Change in potential energy,

$\Delta \text{PE}={\text{mgh}}_f-{\text{mgh}}_i={\text{mg(h}}_f-{\text{h}}_f)$
$\Delta \text{PE}=0.5\times 10\times (0.4-2.1)\text{J}$
$\Delta \text{PE}=-8.5\text{J}$