A body of mass 5kg is initially at rest. When a force of 50N acts on the body placed on the horizontal surface, find the distance it travels in 3 s. Given the coefficient of friction in between the surfaces is 1.

Answers (1)

Friction force acting on the body = μN= μmg = 1 * 5 kg * 9.8=49  N

So, net force acting on the body,

 50 N - friction force =ma

Or, 50 N - 49 N=5a

Or, 1=5a

Or, a=1/5 m/s2

Distance traveled in 3 seconds is s = u t + 1/2 a t² = 0 + 1/2 * 0.2 * 3² = 0.9 m

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