Question

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?

Answer 1

a) frictional force = 93.8 N

b) acceleration = 1.29 ms∧-2

Step-by-step 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