On a planet where g = 10.0 m/s² and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30° as shown in the figure. The object is allowed to move and it stops on a rough horizontal surface, at a distance of 4.0 m from the bottom of the hill. The coefficient of kinetic friction on the hill is 0.40. What is the coefficient of kinetic friction between the horizontal surface and the sled? A) 0.10 B) 0.20 C) 0.31 D) 0.40 E) 0.60

Question

Quizplus · Accepted Answer

We can begin by using conservation of energy. Kinetic energy at the bottom is zero, so the initial potential energy is equal to the final kinetic energy plus the frictional work against the hill and the work against friction on the horizontal surface.
$mgh = \frac{1}{2}mv^2 + \mu_kmgd_1 + \mu_hmgd_2$
where m is the mass of the sled, g is the acceleration due to gravity, h is the height of the hill, d1 is the distance traveled on the hill, d2 is the distance traveled on the horizontal surface, and $\mu_k$ and $\mu_h$ are the coefficients of kinetic friction on the hill and the horizontal surface respectively.
We can solve for $v^2$:
$v^2 = 2gh - 2\mu_kgd_1 - 2\mu_hmgd_2$
When the sled comes to a stop on the horizontal surface, the net force is zero, so the frictional force is equal in magnitude to the component of gravity parallel to the surface:
$\mu_hmg = mgsin(30)$
$\mu_h = sin(30) = 0.5$
Substituting this into the equation for $v^2$, we can solve for $\mu_k$:
$v^2 = 2gh - 2\mu_kgd_1 - 2(0.5)mgd_2$
$0 = 2gh - 2\mu_kgd_1 - 2(0.5)mgd_2$
$\mu_k = \frac{g}{2d_1}(h - 1m)$
Plugging in the given values, we get:
$\mu_k = \frac{10}{2(4)}(3.0 - 1) = 0.31$
Therefore, the answer is choice C.

On a Planet Where G = 10