Question

1. a box of books weighing 320 n is shoved across the floor by a force of 500 n exerted downward at an angle of 30° to the horizontal. what is the coefficient of kinetic friction between the box and the ground if it is moving at a constant speed?

Explanation:

Step1: Analyze vertical forces

The vertical forces on the box are the weight \(W = 320\ N\), the vertical component of the applied force \(F_y=F\sin\theta\) (where \(F = 500\ N\) and \(\theta = 30^{\circ}\)), and the normal force \(N\). Since there is no vertical acceleration (\(a_y = 0\)), by Newton's second law \(\sum F_y=0\). So \(N=W + F\sin\theta\).

$$N=320+500\sin30^{\circ}=320 + 250=570\ N$$

Step2: Analyze horizontal forces

The horizontal forces are the horizontal component of the applied force \(F_x=F\cos\theta\) and the kinetic - friction force \(f_k=\mu_kN\). Since the box is moving at a constant speed (\(a_x = 0\)), by Newton's second law \(\sum F_x=0\). So \(F\cos\theta=f_k=\mu_kN\).
We know \(F = 500\ N\), \(\theta = 30^{\circ}\), and \(N = 570\ N\). Then \(\mu_k=\frac{F\cos\theta}{N}\)

$$F\cos\theta=500\cos30^{\circ}=500\times\frac{\sqrt{3}}{2}\approx433\ N$$

$$ \mu_k=\frac{433}{570}\approx0.76$$

Answer:

The coefficient of kinetic friction \(\mu_k\approx0.76\)