A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surfac

Question

A horizontal force of 80 N acts on a mass of 6 kg resting on a horizontal surface. The mass is initially at rest and covers a distance of 5 m in 0.92 s under the action of the force. Assuming there are no energy losses due to air resistance and therefore that the acceleration is constant: Calculate the total energy expended in the acceleration. Plot a graph of the Kinetic energy of the mass against time Plot a graph of the kinetic energy of the mass against distance Calculate the coefficient of friction between the mass and the surface

Explanation / Answer

If there were no friction, teh force F would do work W given by: W = Fd where d = distance the force acts over so using your numbers W = 80N*5m = 400 J (a) Now it takes the box 0.92s to move the 5m and you are told acceleration is constant so you can use d = 1/2 at^2 to find a ---> a = 2d/t^2 = 2*5 m/(0.92s)^2 = 11.82 m/s^2 The net force, the difference between the applied force F and friction, is then: Fnet = ma = 70.92 =N Then, Fnet = F - umg where u = coefficient of friction and g = 9.8 m/s^2 SOlving for u ---> u = (F - Fnet)/mg = 0.154

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