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

Let's start by finding the velocity of the sled at the bottom of the hill using conservation of energy. The initial potential energy of the sled at the top of the hill is converted to kinetic energy at the bottom of the hill. Since the sled starts from rest, the initial kinetic energy is zero. Thus:

mgh = 1/2 mv^2

Solving for v, we get:

v = sqrt(2gh)

Plugging in the values, we get:

v = sqrt(2*10*6) =sqrt(120) m/s

Now we can use this velocity to find the coefficient of kinetic friction on the horizontal surface. The distance traveled on the horizontal surface is 4 m, and the initial velocity is sqrt(120) m/s. The final velocity is zero since the sled comes to a stop. We can use the kinematic equation:

v^2 = u^2 + 2as

Where u is the initial velocity, a is the acceleration, s is the distance, and v is the final velocity. Solving for the acceleration:

a = (v^2 - u^2)/2s

Plugging in the values, we get:

a = (0 - 120)/2(4) = -15 m/s^2

This acceleration is caused by the force of kinetic friction between the sled and the horizontal surface. Thus:

Ff = ma = -15m

The force of friction on the horizontal surface is:

Ff = N*µk

Where N is the normal force and µk is the coefficient of kinetic friction. We know that the force of gravity is acting downwards, and the normal force is acting perpendicular to the surface. Thus:

N = mg*cos(30) = 8.66m

Plugging in the values, we get:

8.66m*0.40 = -15m

Solving for the coefficient of kinetic friction, we get:

µk = -15/8.66*0.40 = 0.31

Thus, the coefficient of kinetic friction between the horizontal surface and the sled is 0.31. The answer is C.

On a Planet Where G = 10