A 400 N box is being pushed across a level floor at a constant speed by a force P of 100 N at an angle of 30.0° to the horizontal, as shown in the the diagram below. What is the coefficient of kinetic friction between the box and the floor?

A. 0.19

B. 0.24

C. 0.40

D. 0.22

Correct Option: Answer is A

W = 400 N; P = 100 N; θ = 30o; μ = ?

Frictional force (Fr) = μR (where R is the normal reaction)

The forces acting along the horizontal direction are Fr and Px

therefore Pcos 30° – Fr = ma (Pcos 30° is acting in the +ve x-axis while Fr in the -ve x-axis)

⇒ 100cos 30° – μR = ma

Since the box is moving at constant speed, its acceleration is zero

⇒ 100cos 30° – μR = 0

⇒ 100cos 30o = μR —– (i)

The forces acting in the vertical direction are W, Py and R

therefore, R – Psin 30° – W = 0 (R is acting upward (+ve) while Py and W are acting downward (-ve) and they are at equilibrium)

⇒ R – 100sin 30° – 400 = 0

⇒ R = 100sin 30° + 400

⇒ R = 50 + 400 = 450 N

From equation (i)

⇒ 100cos 30° = 450μ

⇒μ=100cos30°

N = 100cos30°/450 = μ = 0.19