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 , determine the kinetic energy of the block at the instant the force ceases to act. Take g = 10 m/s ^2 .

Step-by-step solution

Given the applied force F = 20 N and the work done by this force W = 40 J . The displacement s of the block along the incline can be calculated as: W = Fs 40 = 20s s = 2 m Using the kinematic equation s = ut + 1 2 at^2 with initial velocity u = 0 and time t = 1 s : 2 = 0 + 1 2 a(1)^2 a = 4 m/s ^2 The velocity of the block at t = 1 s is: v = u + at = 0 + 4(1) = 4 m/s The kinetic energy of the block at this instant is: K = 1 2 mv^2 = 1 2 (2.0)(4)^2 = 16 J Alternatively, using the work-energy theorem: The component of