Question

A crate of mass 20 kg is being pushed with a horizontal force of 63 N, moving with a constant velocity. A. Find the co-efficient of kinetic friction.B. If a friend of mass 60 kg sits on the crate , find the acceleration

Answer 1

Given data

*The given mass of the crate is m = 20 kg

*The crate is pushed with a horizontal force is F = 63 N

*The mass of the friend is M = 60 kg

(A)

As the block is moving with constant velocity, it means zero net force acting on the crate. Hence, the frictional force is f = 63 N

The formula for the coefficient of kinetic friction is given as

`f=\mu_kmg`

Substitute the known values in the above expression as

`\begin{gathered} 63=\mu_k*(20)(9.8) \\ \mu_k=0.32 \end{gathered}`

(B)

The formula for the acceleration is given as

`a=(F)/((m+M))`

Substitute the known values in the above expression as

`\begin{gathered} a=(63)/((20+60)) \\ =0.78m/s^2 \end{gathered}`