Question

(problem reference 4 - 2) a boy pulls a 5.00 - 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? a. 0.24 b. 0.06 c. 0.09 d. 0.18 e. 0.12

Explanation:

Step1: Analyze vertical - force equilibrium

The vertical forces acting on the sled are the gravitational force $F_g = mg$, the vertical component of the pulling force $F_{y}=F\sin\theta$, and the normal force $N$. Since there is no acceleration in the vertical direction ($a_y = 0$), according to Newton's second - law $\sum F_y=0$. So $N + F\sin\theta=mg$, and $N=mg - F\sin\theta$. Given $m = 5.00\ kg$, $g = 9.8\ m/s^2$, $F = 10.0\ N$, and $\theta = 60^{\circ}$, we have $N=5\times9.8-10\times\sin60^{\circ}=49 - 10\times\frac{\sqrt{3}}{2}=49 - 5\sqrt{3}\ N$.

Step2: Analyze horizontal - force equilibrium

The horizontal forces acting on the sled are the horizontal component of the pulling force $F_{x}=F\cos\theta$ and the kinetic friction force $f_k=\mu_k N$. Since the sled moves with constant velocity ($a_x = 0$), according to Newton's second - law $\sum F_x = 0$. So $F\cos\theta=f_k=\mu_k N$. Then $\mu_k=\frac{F\cos\theta}{N}$. Substitute $F = 10\ N$, $\theta = 60^{\circ}$, and $N=49 - 5\sqrt{3}\ N$ into the formula. $F\cos\theta=10\times\cos60^{\circ}=5\ N$, $N = 49-5\sqrt{3}\approx49 - 5\times1.732=49 - 8.66 = 40.34\ N$. So $\mu_k=\frac{5}{40.34}\approx0.12$.

Answer:

e. 0.12