A person pushes horizontally with a force of 180. N on a 67.0 kg crate to move it across a level floor. The coefficient of kinetic friction is 0.14. (a) What is the magnitude of the frictional force? (b) What is the magnitude of the crate's acceleration?

a) frictional force = 93.8 N

b) acceleration = 1.29 ms∧-2

Explanation:

The frictional force can be calculated by:

μ k = Fk / N

where:

μ k = coefficient of kinetic friction = 0.14 N = normal force or the force perpendicular to the contacting surface Weight of the crate, W = mg = (67 * 10) = = 670 N

take acceleration due to gravity = 10m/s∧2

The normal force, N = W = 670N (upward force equals downward force)

Fk = μ k * N

= 0.14 * 670

Fk = 93.8 N

b) To calculate the magnitude of crate's acceleration

F - Fk = ma

where

Fk = frictional force = 93.8N F = horizontal force = 180 N m = mass = 67.0 kg a = magnitude of acceleration (unknown)

by substituting,

180 - 93.8 = 67a

86.2 = 67a

acceleration, a = (86.2 : 67)

a = 1.29 ms∧-2