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 frictional force acting on the block during the first second. Take g = 10 m/s ^2 .

Step-by-step solution

The displacement of the block in the first second is given by the equation of motion: 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 forces acting on the block parallel to the inclined plane are the applied force F , the component of gravity mg , and the frictional force f acting opposite to the motion. The equation of motion is: F + mg - f = ma Substituting F = 20 N , m = 2.0 kg , g = 10 m/s ^2 , = 37^ , and a =