A block of mass m = 10 kg is released from the top of the smooth inclined surface of a wedge which is moving horizontally toward right at a constant velocity of u = 10 m/s. Inclination of the wedge is `theta=37^(@)`. Calculate the work done by the force applied by the wedge on the block in two seconds in a reference frame attached to -
(a) the ground (b) the wedge.
[Take `g=10 m//s^(2)`]

Answer

(a) 960 J
(b) zero