A block of mass 2.0 kg initially at rest on an inclined plane of inclination 37^ is pulled up the plane by a constant force of 20 N applied parallel to the incline. The force acts for one second. If the work done by the applied force is 40 J , compute the work done by the force of gravity in that one second. Take g = 10 m/s ^2 .

Step-by-step solution

Given the work done by the applied force W_F = 40 J and the applied force F = 20 N . The displacement of the block along the incline is given by: s = W_F F = 40 20 = 2 m The block moves 2 m up the incline. The vertical displacement of the block is h = s 37^ . The work done by the force of gravity is negative since the displacement is upwards while gravity acts downwards: W_g = -mgh = -mgs 37^ Substituting the given values ( m = 2.0 kg , g = 10 m/s ^2 , 37^ = 3 5 ): W_g = -2.0 10 2 3 5 W_g = -12 2 = -24 J