Question

1. a 1.95 kg box sits on an incline of 24° with the horizontal. if the box accelerates down the incline at 0.245 m/s², what is the coefficient of kinetic friction between the box and the inclined plane? 0.42 0.26 0.62 0.31

Explanation:

Step1: Resolve forces

The gravitational force along the incline is \(mg\sin\theta\), the normal force is \(N = mg\cos\theta\), and the frictional force is \(f=\mu_kN=\mu_kmg\cos\theta\). Using Newton's second law \(F = ma\), we have \(mg\sin\theta-\mu_kmg\cos\theta=ma\).

Step2: Solve for \(\mu_k\)

Divide the equation \(mg\sin\theta-\mu_kmg\cos\theta=ma\) by \(mg\cos\theta\). We get \(\tan\theta-\mu_k=\frac{a}{g\cos\theta}\). Then \(\mu_k=\tan\theta-\frac{a}{g\cos\theta}\).
Substitute \(m = 1.95\space kg\), \(\theta = 24^{\circ}\), \(a=0.245\space m/s^{2}\), \(g = 9.8\space m/s^{2}\)
\(\tan(24^{\circ})\approx0.445\), \(\cos(24^{\circ})\approx0.914\), \(\frac{a}{g\cos\theta}=\frac{0.245}{9.8\times0.914}\approx0.027\)
\(\mu_k=0.445 - 0.027=0.418\approx0.42\)

Answer:

0.42