Question

(problem reference m.6)
a boy pulls a 5.0 - kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. the boy pulls on the rope with a force of 10.0 n, and the sled moves with constant velocity.
what is the coefficient of kinetic friction between the sled and the ice?
0.18
0.12
0.09
0.06
0.24

Explanation:

Step1: Analyze forces in x - direction

Since the sled moves with constant velocity, \(F_{net,x}=0\). The horizontal component of the pulling force \(F\cos\theta\) equals the frictional force \(f\). So \(f = F\cos\theta\), where \(F = 10.0\ N\) and \(\theta=60.0^{\circ}\), then \(f=10\cos60^{\circ}=10\times0.5 = 5\ N\).

Step2: Analyze forces in y - direction

\(F_{net,y}=0\). The normal force \(N\) plus the vertical component of the pulling force \(F\sin\theta\) equals the weight \(mg\). So \(N=mg - F\sin\theta\), where \(m = 5.0\ kg\), \(g = 9.8\ m/s^{2}\), \(F = 10.0\ N\) and \(\theta = 60.0^{\circ}\). Then \(N=5\times9.8-10\sin60^{\circ}=49 - 10\times\frac{\sqrt{3}}{2}=49 - 5\sqrt{3}\approx49 - 8.66=40.34\ N\).

Step3: Calculate the coefficient of kinetic friction

The formula for kinetic friction is \(f=\mu_{k}N\). Rearranging for \(\mu_{k}\), we get \(\mu_{k}=\frac{f}{N}\). Substituting \(f = 5\ N\) and \(N\approx40.34\ N\), \(\mu_{k}=\frac{5}{40.34}\approx0.12\).

Answer:

0.12