Question

A 12.0 kg box is being pulled along the level ground at a constant velocity by a horizontal force of 38.0 N. What is the coefficient of kinetic friction between the box and the floor?

Answer 1

Explanation:

Hello mate it's easy...

Force is applied that means the box must move with ACCELERATION. But here in question box is moving Constant velocity that means the 38 N which is being applied is cancelled by friction as a result it's moving with constant Velocity.

Now moving towards simple Mathematics.

we know Frictional Force = ßmg

ß = Co-efficient of Friction

m = mass of Object

g = ACCELERATION due to gravity

now we know , friction cancels the 38 N force

thus.

38 - ßmg = 0

38 = ßmg

38 = ß (12)(9.8)

ß = [38/(12)(9.8)]

ß = 0.323 ~ 0.3

Thus Coefficient of Friction ß = 0.3