Question

What is the work done by moving a body of mass 60 Kg through a distance of 5 metres
along the horizontal plane, if the coefficient friction is 0.2?
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Answer 1

Answer:m = 2 kg                 F = 20 N                   t = 10 s

Friction force = μ m g = 0.2 * 2 * 10 = 4 N

Net force in forward direction = 20 - 4 = 16 N

So acceleration = a = Force / m  = 16 /2 = 8 m/s²

Initial speed = u = 0

Distance traveled = s = u t + 1/2 a t² = 0 + 1/2 * 8 * 10² = 400 m

Work done by F = F * s = 20 * 400 = 8 kJ

Work done by Friction force = - 4 * 400 = - 1.6 kJ  (heat dissipated)

Kinetic energy of the body = 1/2 m v² = 1/2 m (2 a s)

      = 2 * 8 * 400 = 6.4 kJ

Solve same as this

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