A horizontal force of 150 N is used to push a 38.0 kg packing crate a distance of 6.85 m on a rough horizontal surface. If the crate moves at constant speed, find each of the following. a) find the work done by the 150N force:
b) find the coefficient of kinetic energy between the crate and rough surface

0.40

Explanation:

a) Work done is expressed as

W = Force × distance

Given.

Force = 150N

Distance = 6.85m

Workdone = 150×6.85

Work done = 1027.5Joules

b) According to Newton's second law

\sumFx = ma

Fm - Ff = ma

Since speed is constant, acceleration us zero

Fm - Ff = 0

Fm = Ff = nR

Fm is the moving force

Ff is the frictional force

n is the coefficient of kinetic energy between the crate and rough surface

R is the reaction

From the formula;

Fm = nR

n = Fm/R

n = Fm/mg

n = 150/38(9.8)

n = 150/372.4

n = 0.40

Hence the coefficient of kinetic energy between the crate and rough surface is 0.40

are there answer choices to this question? this seems like a question where you have options to choose from.