Question

A 102-kg crate is being pushed across a horizontal floor by a force P that makes an angle of 41.7 ° below the horizontal. The coefficient of kinetic friction is 0.232. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

Answer 1

The force that causes the crate to accelerate is the horizontal component of P. The force that causes the crate to decelerate is the kinetic frictional force. Since the net work is zero, these two forces are equal in magnitude. There are two forces that are perpendicular to the floor. These are the weight of the crate and the vertical component of P. Since the direction of P is below horizontal, the total vertical force is equal to the sum of these two forces.

Horizontal component = P * cos 41.7
Vertical component = P * sin 41.7
Weight = 102 * 9.8 = 999.6 N

Total vertical force = P * sin 41.7 + 999.6
Ff = 0.232 * (P * sin 41.7 + 999.6 )
Ff = 0.232 * P * sin 41.7 + 231.9
Set these two forces equal to each other and solve for P.

P * cos41.7 = 0.232 * P * sin 41.7 + 231.9
P * (cos41.7 – 0.232* sin41.7) = 231.9
P =231.9/(cos41.7 – 0.232* sin41.7)

P is approximately 391.5 N.

I hope this is helpful for you.