Law of Conservation of Energy of a skier

[tongtong]

A skier of mass 55.0 kg slides down a slope 11.7 m long, inclined at an angle theta to the horizontal. The magnitude of the kinetic friction is 41.5 N. The skier's initial speed is 65.7 cm/s and the speed at the obttom of the slope is 7.19 m/s. Determine the angle theta from the law of conservation of energy. Air reistance is negligible.

The answer is 17.5 degrees, but the work is hard for me to start at.
From what I've learned, we use Et1 = Et2, and I can find theta from 11.7 sin theta which is delta y. I get 12.9 degrees from this. I know this is incorrect, and the friction force needs to be added in somewhere, I just don't know where.

 

[Doc Al]

Originally posted by tongtong
I know this is incorrect, and the friction force needs to be added in somewhere, I just don't know where.

If it wasn't for friction, the mechanical energy (KE + PE) would be conserved. The change in KE + PE (a decrease, of course) is equal to the work done by friction on the skier. (Don't forget that work is negative if the force is opposite the displacement.)

 

[M98Ranger]

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred from one form to another. In the case of the skier sliding down the slope, the initial potential energy at the top of the slope is converted into kinetic energy as the skier accelerates down the slope. This kinetic energy is then converted back into potential energy as the skier reaches the bottom of the slope.

To determine the angle theta from the Law of Conservation of Energy, we can use the equation Et1 = Et2, where Et1 is the initial total energy (potential + kinetic) and Et2 is the final total energy (potential + kinetic). In this case, the initial energy is purely potential energy at the top of the slope, and the final energy is a combination of potential and kinetic energy at the bottom of the slope.

Et1 = mgh = (55.0 kg)(9.8 m/s^2)(11.7 m) = 6452.1 J
Et2 = (1/2)mv^2 + mgh = (1/2)(55.0 kg)(7.19 m/s)^2 + (55.0 kg)(9.8 m/s^2)(0 m) = 2285.4 J

Setting Et1 equal to Et2, we can solve for the angle theta:

Et1 = Et2
mgh = (1/2)mv^2 + mgh
mgh - mgh = (1/2)mv^2
0 = (1/2)mv^2
0 = (1/2)(55.0 kg)(7.19 m/s)^2
0 = 2285.4 J

Solving for theta, we get:

theta = sin^-1 (v^2/2gh)
theta = sin^-1 [(7.19 m/s)^2/(2)(9.8 m/s^2)(11.7 m)]
theta = 17.5 degrees

Therefore, the angle theta from the Law of Conservation of Energy is 17.5 degrees. This accounts for the effect of friction, as it is included in the initial potential energy at the top of the slope.