5 Part Physics Question - Could somebody please help, please explain your answers to me, it would help me a lot

Question

Caleb, an honor Chemistry student, went a ski trip with his friends to the Widham Ski Resort in

NY. The careful Caleb chose one of blue routes to do downhill skiing, which is 30o slope, 100 m

high, and has the coefficient of kinetic friction of 0.1. If he, 70 kg, starts from rest at the top,

(a) Draw a free‐body diagram for Caleb.

(b) what is his speed at the bottom of the slope?

(c) What is his acceleration during downhill skiing assuming the acceleration is constant?

(d) At the bottom of the slope, he encounters a horizontal surface with the coefficient of kinetic

friction of 0.5. What’s his acceleration on the horizontal surface when he comes to rest?

(e) How far does he travel before coming to rest?

Answer 1

a. I can't draw the free body diagram, but it consists of a downward force due to gravity

W = mg = 70 kg x 9.8 m/s^2 = 686 N. There is also a force normal to the slope

N = mgcos30 = 70kg x 9.8 m/s^2 x 0.866 = 594.1 N. There is another force parallel to the slope pointing down the slope P = mgsin30 = 343 N. Lastly. there is a frictional force parallel to the slope but pointing up the slope Fr = μ x N = 0.1 x 594.1 = 59.41 N.

b. The potential energy when the skier is at the top of the slope is

PE = mgh = 70kg x 9.8 m/s^2 x 100 m = 68600 J.

This potential energy must be equal to the kinetic energy at the bottom of the slope plus the energy expended overcoming the frictional force.

PE = KE + Fr x D where D is the length of the slope which is 100 m/sin30 = 200 m.

KE = PE - Fr x 200 = 68600 J- 11882 J = 56718 J. This is equal to the KE or 1/2 x m x V^2

Solving for V = √ (2 x 56718)/70 = 40.3 m/s

c. To find the acceleration down the slope, we can use the equation V^2 = Vo^2 + 2aD and solve for a

Vo is 0, so V^2 = 2aD, a = V^2/2D = 40.3^2/400 = 4.06 m/s

d/e. Once on the horizontal portion, the KE at the start must equal the energy bringing the skier to a stop which is the work done overcoming friction. The frictional work is 0.5 x 70 kg x 9.8 m/s^2 x S where S is the distance travelled before the stop. Set up the equation

1/2 x70 x 40.3^2 = 0.5 x 70 x 9.8 x S. Solve for S = 166 m/s

To get the deceleration, use Vf^2 = Vo^2 - 2aS Vf is 0, so a = 40.3^2/-2x 166 = - 4.89 m/s^2

I urge you to check the math and reasoning. I left out units in some of the calculation to save time and space.

Answer 2

Something I forgot to say when I derived that last equation for the distance that it took to come to rest, the µ in THAT denominator specifically is the SECOND µ that is talked about in your problem prompt, not the original. Only realized after I finished recording my answer I hadn't said that.

I hope this helps you!