Question

How much force must be applied to push a 253.2 kg crate across the floor at a constant velocity if the coefficient of kinetic friction is 0.55?

Answer 1

Given,

The mass of the crate, m= 253.2 kg

The coefficient of the kinetic friction between the floor and the crate, μ=0.55

Given that the crate is pushed with a constant velocity. That is the net force on the crate is zero.

The only two forces acting on the crate are the force with which it is being pushed and the friction that is opposing the applied force.

The net force on the crate is given by,

`\begin{gathered} F_n=0=F-f \\ =F-mg\mu \end{gathered}`

Where f is the frictional force between the floor and the crate, F is the applied force, and g is the acceleration due to gravity.

On substituting the known values,

`\begin{gathered} F-253.2*9.8*0.55=0 \\ \Rightarrow F=253.2*9.8*0.55 \\ =1364.75\text{ N} \end{gathered}`

Thus the force with which the crate must be pushed is 1364.75 N