Question

1. demarius has just moved, and is unloading boxes the moving van. he slides a 19.0 kg box down the 31° ramp. the box accelerates at a rate of 0.68 m/s². what is the coefficient of kinetic friction between the box and the ramp? 0.61 0.77 0.52 0.59

Explanation:

Step1: Analyze forces along the ramp

The force of gravity along the ramp is \(F_{g\parallel}=mg\sin\theta\), the frictional force is \(F_f = \mu_kmg\cos\theta\). According to Newton's second law \(F_{net}=ma\), so \(mg\sin\theta-\mu_kmg\cos\theta=ma\).

Step2: Solve for \(\mu_k\)

First, 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 \(\mu_k=\tan\theta-\frac{a}{g\cos\theta}\). Given \(m = 19.0\space kg\), \(\theta = 31^{\circ}\), \(a=0.68\space m/s^2\), \(g = 9.8\space m/s^2\).
\(\tan31^{\circ}\approx0.60\), \(\cos31^{\circ}\approx0.86\), \(\frac{a}{g\cos\theta}=\frac{0.68}{9.8\times0.86}\approx0.08\)
\(\mu_k=0.60 - 0.08=0.52\)

Answer:

0.52