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

Answer:

The value is $F_f = 46.935 \ N$

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  

            $92.7 - F_f = 40.5 * 1.13$

=>         $F_f = 46.935 \ N$