Question

51. A 2 g ball of glass is released from the edge of a
hemispherical cup whose radius is 20 cm. How
much work is done on the ball by the gravitational
force during the ball's motion to the bottom of the
cup ?
(1) 1.96 m)
(2) 3.92 m)
modpalopata how in
(3) 4.90 md
isoris
(4) 5.88 m)​

Answer 1

hope it helps !

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