Question

1. a) calculate the acceleration of a skier heading down a 10.0° slope, assuming the coefficient of friction for waxed wood on wet snow. b) find the angle of the slope down which this skier could coast at a constant velocity. you can neglect air resistance in both parts, and you will find the result of exercise 5.9 to be useful. explicitly show how you follow the steps in the problem - solving strategies.

Explanation:

Step1: Identify forces acting on skier

The forces acting on the skier on the slope are the gravitational - force component along the slope $F_{g\parallel}=mg\sin\theta$ and the frictional force $F_f = \mu_kmg\cos\theta$. According to Newton's second - law $F_{net}=ma$. So, $mg\sin\theta-\mu_kmg\cos\theta = ma$.

Step2: Solve for acceleration

We can factor out $mg$ from the left - hand side of the equation: $mg(\sin\theta-\mu_k\cos\theta)=ma$. Then, cancel out the mass $m$ from both sides of the equation. We get $a = g(\sin\theta-\mu_k\cos\theta)$. Given $\theta = 10.0^{\circ}$ and assume a typical value of $\mu_k$ for waxed wood on wet snow, say $\mu_k = 0.1$. And $g = 9.8\ m/s^2$. Then $\sin(10.0^{\circ})\approx0.174$ and $\cos(10.0^{\circ})\approx0.985$. So, $a=9.8(0.174 - 0.1\times0.985)=9.8(0.174 - 0.0985)=9.8\times0.0755 = 0.74\ m/s^2$.

Step3: For part (b)

When the skier moves at a constant velocity, the net force $F_{net}=0$. So, $mg\sin\theta-\mu_kmg\cos\theta = 0$. Then $mg\sin\theta=\mu_kmg\cos\theta$. We can cancel out $mg$ from both sides and get $\tan\theta=\mu_k$.

Answer:

a) The acceleration $a = 0.74\ m/s^2$.
b) The angle $\theta=\arctan(\mu_k)$. If we assume $\mu_k = 0.1$, then $\theta=\arctan(0.1)\approx5.7^{\circ}$