Question

iv) Find the final velocity of a 15.0 kg box if it starts from rest and it slides freely (with no applied forces) for 1.89 meters down an inclined plane that makes a 16.2° angle with the horizontal and the coefficient of friction is 0.121

Answer 1

F = mu N_0 = mu mg cos theta equation of motion is, mg sin - f = ma or, mg sin theta - mu mg cos theta = ma or, a = g(sin theta - mu cos theta), theta = 16.2 degree, mu = 0.121 so, a = 1.59 m/s^2 v^2_f = v^2_i + 2 as = 0^2 + 2a (1.889 m) or, v_f = 2.46 m/s

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