A block of mass 2.0 kg is pushed down an inclined plane of inclination 37^ by a 20 N force acting parallel to the incline. It is observed that the block travels on the incline with an acceleration of 10 m/s ^2 . Assuming the block starts from rest, determine the work done by the weight of the block during the first second. Take g = 10 m/s ^2 .

Step-by-step solution

The distance traveled by the block in the first second is given by the kinematic equation: s = ut + 1 2 at^2 Since the block starts from rest, u = 0 . Substituting the given values a = 10 m/s ^2 and t = 1 s : s = 0 + 1 2 (10)(1)^2 = 5 m The weight of the block acts vertically downwards, and its component parallel to the inclined plane is mg . The work done by the weight of the block is the product of this parallel component and the displacement along the incline: W = (mg ) s Substituting m = 2.0 kg , g = 10 m/s ^2