## 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 diagram below. What is the coefficient of kinetic friction between the box and the floor?

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

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

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