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 sled 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 start by finding the initial potential energy of the sled at the top of the hill:
PE = mgh = (m)(10)(sin 30°)(4) = 20m

Since the sled comes to a stop, all of this potential energy is converted to kinetic energy at the bottom of the hill:
KE = 1/2 mv^2 = 20m

Now we can use work-energy principle to find the force of friction acting on the sled on the horizontal surface:
W_friction = KE_final - KE_initial = -20m
W_friction = F_friction * d = (µ_k)(mg)(d)
-20m = (µ_k)(mg)(4)
µ_k = -5/2g

We know that the coefficient of kinetic friction cannot be negative, so we take the absolute value:
µ_k = 5/2g ≈ 0.31

Therefore, the coefficient of kinetic friction between the horizontal surface and the sled is approximately 0.31, which is answer choice C.

On a Planet Where G = 10