How Far Must a Skier Travel to Reach 56 m/s on a 30° Slope?

[7at1blow]

Homework Statement

A skier reaches a speed of 56 m/s on a 30° slope w/ no friction. What is the minimum distance the skier would have to travel if starting from rest.

Homework Equations

Vf² = Vi² +2(a)(Δd)

The Attempt at a Solution

I hope no one will scoff too hard, this is my first physics class. I made up a mass of 10kg and calculated the weight in x direction with W_x = W(sin(30))= 98.1N, normal force in x is zero.

Then I found weight in y direction with W_y = W(cos(30))= -85.0N

From there I used ƩF_x=M(a_x) and came out with a_x=4.91 m/s² and came to a final solution of ≈320m after filling in the kin. equation.

I spent over an hour on this and feel like I may have just made up nonsense and then "math'd" it. Could someone kindly check my work and give me some guidance?

Thanks.

 

[anathema]

Hello there,

First of all, congratulations on taking your first physics class! It can be a challenging subject, but with practice and perseverance, you will definitely improve.

Your approach to the problem seems to be in the right direction. Let's go through it step by step:

1. The first thing you did was to calculate the weight of the skier in the x and y directions. This is correct, as we need to consider the forces acting on the skier in both directions.

2. Next, you used the equation ƩFx=M(ax) to find the acceleration of the skier in the x direction. This is also correct, as we know that the only force acting on the skier in the x direction is the component of the weight, and this force is responsible for the acceleration.

3. Finally, you used the kinematic equation Vf2 = Vi2 +2(a)(Δd) to find the distance traveled by the skier. This is also correct, as we know the final velocity (56 m/s), initial velocity (0 m/s), and acceleration (4.91 m/s2) of the skier.

Therefore, your solution of approximately 320m seems to be correct. However, there are a few points that you could improve upon:

1. The mass of the skier is not given in the problem, so you don't need to assume a mass of 10kg. This won't affect your final answer, but it's good to get into the habit of not assuming values that are not given.

2. When finding the weight in the y direction, you used Wy = W(cos(30))= -85.0N. This is not correct. The weight in the y direction is equal to the normal force, which is perpendicular to the slope. Therefore, the weight in the y direction should be equal to W(cos(30))= 85.0N.

3. Lastly, it would be good to mention the units in your final answer. Since you used meters for distance and seconds for time, the unit for acceleration would be m/s2, and the unit for final answer would be meters.

Overall, your approach to the problem was good, and you arrived at the correct answer. Keep practicing and you will definitely improve. Good luck!