Question

A force of 400-N pushes on a 25-kg box horizontally. The box accelerates at 9 m/s? Find the coefficient of kinetic friction between the box and floor.

Answer 1

Answer:

0.69

Explanation:

Using the Newtons law of motion;

$\sum Fx = ma_x\\Fm - Ff = ma_x$

Fm is the moving force = 400N

Ff is the frictional force = μR

μ is the coefficient of kinetic friction

R is the reaction = mg

m is the mass

a is the acceleration

The equation becomes;

$Fm - \mu R = ma_x\\Fm - \mu mg = ma_x\\400- \mu (25)(9.8) = 25(9)\\400 - 254.8 \mu = 225\\- 254.8 \mu = 225 - 400\\- 254.8 \mu = -175\\ \mu = \frac{-175}{- 254.8} \\\mu = 0.69$

Hence the coefficient of kinetic friction between the box and floor is 0.69