Question

Some sliding rocks approach the base of a hill with a speed of 17.0 m/s . The hill rises at 42.0 ∘ above the horizontal and has coefficients of kinetic and static friction of 0.450 and 0.550, respectively, with these rocks. Start each part of your solution to this problem with a free-body diagram. Find the acceleration of the rocks as they slide up the hill. Once a rock reaches its highest point, will it stay there or slide down the hill? If it slides down, find its acceleration on the way down, else enter 0

Answer 1

Notice that the mass of the rock is not given. So what we are going to do is to get the mass out the rocket canceled.

First steps, draw a free body diagram to help you visualize.
secondly, the speed of the ball is just an extra information, no matter how fast it goes, acceleration is still the same.

Find the weight of the rock
Weight = mg

When the rock is on the ramp, part of its weight is taken down or parrellel the the ramp (x-direction) and part of the weigh is taken perpendicular to the ramp (y-direction)

Break down the weight
weight(x) = sin(32)mg
weight(y) = cos(32)mg

The Fnormal is the force that is perpendicular to the ramp, weight(y) is also perpendicular to the ramp. According the Newton's 3rd law, both the weight(y) and the normal force are equal

Fnormal = cos(42)mg

Friction = μ(Fnormal)
plug cos(42) for Fnormal
Friction = μ (cos(42)mg )
Fnet = ma

Fnet is the sum of forces. In this case, both weight(x) and Friction make up the net force. Since both weight(x) and Friction are in the same direction, the net force is the sum of the two forces

Fweight(x) + Friction = ma
sin(42)mg + μ (cos(42)mg = ma
Now the mass cancel out. you left with
sin(42)g + μ(cos(42)g = a
plug 9.8 for g and 0.45 for μ
sin(42)9.8 + 0.45*cos(42)9.8 = a
a = 9.385 m/s (should be -9.385 m/s^2 because the rocks slows down as it goes)


b) Now that the rocks slides downs, firction causes the rocks to accelerate less than it slides up. Now that the friction force is opposite of the weight (x). Weight(x) - Friction = ma

sin(42)mg - μ(cos(42)mg) = ma
sin(42)g - μ(cos(42)g = a
a = sin(42)9.8 – 0.45(cos(42)9.8)
a = 3.28 m/s^2