Question

A 2 g ball of glass is released from the edge of a hemispherical cup whose radius is 20cm . How much work is done on the ball by the gravitational force during the ball's motion to the bottom of the cup ? Gxuvxgnm

Answer 1

As the Gravitational force is CONSERVATIVE in nature hence Work done by it depends only on initial and find position.
_____________
We will consider only VERTICAL direction only. Because we have to take that displacement which is along the Gravitational force .


______________
Initially the ball at the top of the hemispherical cup whose radius is 20cm.it means that initially the ball at height of 20cm from the ground .
And finally it comes to ground .
=> Displacement of ball = 20 cm
= 0.20 m .
_____________
Given that mass (m) of ball is 2g = 0.002Kg.
So Gravitational Force acting on it = mg
= 0.002×10 = 0.02N.
______________
Finally WORK DONE = force ×displacement ;
= 0.02×0.20 = 0.004 J
_______________
hope it helps !