Question

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, what is the magnitude of the force of kinetic friction (in N) acting on the crate

Answer 1

The value is
`F_f = 46.935 \ N`

Step-by-step explanation:

From the question we are told that

The magnitude of the horizontal force is
`F = 92.7 \ N`

The mass of the crate is
`m = 40.5 \ kg`

The acceleration of the crate is
`a = 1.13 \ m/s`

Generally the net force acting on the crate is mathematically represented as

`F_(net) = F - F_f = ma`

Here
`F_f` is force of kinetic friction (in N) acting on the crate

So

`92.7 - F_f = 40.5 * 1.13`

=>
`F_f = 46.935 \ N`