9702/May Jun/12/2015/Q15

A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests on the edge
of a step of depth 0.8 m, as shown.

The block is knocked over the edge of the step and rotates through 90° before coming to rest with

the 0.6 m edge horizontal.

What is the change in gravitational potential energy of the block?

A 300 J 

B 400 J 

C 450 J 

D 550 J

Solution:

Answer: C.

Since the block is said to be uniform, the centre of mass can be considered to be at the centre.

At the top of the step,

Height of position of centre of mass above ground = 0.8 + (0.6/2) = 0.8 + 0.3 = 1.1 m

At the bottom of the step,

Height of position of centre of mass above ground = 0.4/2 = 0.2 m

Change in GPE = mgΔh = 500 × (1.1 – 0.2) = 450 J

Reference: PYQ - May/Jun 2015 Paper 12 Q15