Energy-related problem involving skiing up a hill

[Emethyst]

Homework Statement

A 68kg skier approaches the foot of a hill with a speed of 15m/s. The surface of the hill slopes up at 40 degrees above the horizontal and has the coefficients of static and kinetic friction of 0.75 and 0.25, respectively. Use energy conservation to find the maximum height above the foot of the hill that the skier will reach.

Homework Equations

Conservation of energy, work-energy theorem

The Attempt at a Solution

I have no idea how to solve this question; since there is friction present in the system I decided to use the work-energy theorem, thus making: E_f = E_i + W, or mgh = (1/2)mv^2 + W. The problem I'm running into is trying to figure out how to calculate a value for W seeming as there are no displacement values given in the question. How would I go about getting an answer for W, or is this the wrong way to go about this question? Any help would be greatly appreciated, thanks in advance.

 

[kuruman]

Call the height that you are looking for y. The displacement along the hillside (for purposes of calculating work) is s = y / sin40^o. So if you use the work energy theorem, you will have an equation involving only one unknown, y.

 

[kerimek]

Hello,

Thank you for sharing your attempt at solving the energy-related problem involving skiing up a hill. It is great that you have recognized the presence of friction and have decided to use the work-energy theorem to approach this question. However, you are right in saying that there is a missing value for W, which represents the work done by friction.

To solve this problem, we can use the concept of conservation of energy. This states that the total energy of a system remains constant, meaning the initial energy (Ei) is equal to the final energy (Ef). In this case, the initial energy is the kinetic energy (1/2)mv^2 of the skier, and the final energy is the potential energy (mgh) at the maximum height reached by the skier.

We can set up an equation using these values: (1/2)mv^2 = mgh. We can cancel out the mass (m) on both sides of the equation, leaving us with v^2 = 2gh. We can then solve for h by substituting the given values for v (15m/s) and g (9.8m/s^2), giving us h = 11.47m.

However, this is not the final answer because we need to take into account the work done by friction. We can calculate this by multiplying the coefficient of kinetic friction (0.25) by the normal force (mgcosθ) and the displacement (h/sinθ) of the skier up the hill, giving us: W = 0.25mgcosθ(h/sinθ).

Substituting the values, we get W = 1.74mg. We can then add this value to the potential energy at the maximum height (mgh) to get the final energy, which is equal to the initial kinetic energy. We can set up an equation: (1/2)mv^2 = mgh + W. Substituting the values, we get: (1/2)(68)(15)^2 = (68)(9.8)(11.47) + 1.74(68)(9.8). Solving for h, we get h = 11.71m.

Therefore, the maximum height above the foot of the hill that the skier will reach is approximately 11.71m. I hope this helps in your understanding of energy conservation and solving problems involving friction.