A 50 kg box is at rest on a horizontal surface. When an 80 N force is applied to the box, it slides on the surface with a constant velocity. How will the force of friction change if an identical box is placed on top of the 50 kg box

Question

Israel Castaneda

Physics

29 January, 16:14

A 50 kg box is at rest on a horizontal surface. When an 80 N force is applied to the box, it slides on the surface with a constant velocity. How will the force of friction change if an identical box is placed on top of the 50 kg box

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Karter · Answer 1

As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.

Then friction force = 80 Newtons but in the opposite direction.

Friction force = Mu * Normal force exerted by ground = Mu * weight of box

So we find Mu.

Mu = coefficient of friction between box and horizontal surface

= Force of friction / weight = 80 / 50 * 9.81 = 0.163

When an identical box is placed on top, the force of friction is

= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons