Energy calculations for a skier on a hill

AI Thread Summary
The discussion focuses on calculating the gravitational potential energy of a skier at the top of a 45 m hill and the kinetic energy at the bottom with a speed of 7.2 m/s. It highlights that the gravitational potential energy is not fully converted into kinetic energy due to factors like friction and air resistance, which dissipate energy. The choice of the zero point for potential energy calculations is arbitrary, but it is suggested that the bottom of the hill is a reasonable reference point. The conversation emphasizes the importance of understanding energy conversion and the effects of external forces in such calculations. Overall, the complexities of energy dynamics in skiing are acknowledged.
Meeeessttteeehh
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Thread title changed to be more descriptive of the problem.

Homework Statement


A skier with a mass of 45 kg is standing at the top of a 45 m hill.
· Calculate the gravitational potential energy of the skier when she is standing at the top of the hill
· Calculate the kinetic energy of this skier at the bottom of the hill, where she has a speed of 7.2 m/s
· Explain why the kinetic energy of the skier at the bottom of the hill is not equal to the gravitational potential energy of the skier at the top of the hill

Homework Equations


E_g =mgh
E_K=1/2mv^2

The Attempt at a Solution


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Looks very good, but is it true that the majority of the initial gravitational potential energy is converted into kinetic energy? Also, does your reference to "friction" include air resistance?
 
Your answer is fine. My only observation is that the problem did not specify where the origin for calculating the potential energy is to be taken. This point is arbitrary. For example if the zero point of potential energy were at the top of the hill, then its value would be zero at the top and - 2×104 J at the bottom. However, the choice of zero does not change the fact that when the skier reaches the bottom of the hill the speed will be the same. It's the difference in potential energy that counts.
 
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TSny said:
Looks very good, but is it true that the majority of the initial gravitational potential energy is converted into kinetic energy? Also, does your reference to "friction" include air resistance?
Thanks! No to the air resistance, I forgot about that... should I add it in? Or would it be "out of place" as I was taking about energy and air resistance isn't a type of energy? As for the conversion, the numbers suggest that the majority is converted to kinetic, but I guess this requires more research... Who knew skiing was so complicated!
 
kuruman said:
Your answer is fine. My only observation is that the problem did not specify where the origin for calculating the potential energy is to be taken. This point is arbitrary. For example if the zero point of potential energy were at the top of the hill, then its value would be zero at the top and - 2×104 J at the bottom. However, the choice of zero does not change the fact that when the skier reaches the bottom of the hill the speed will be the same. It's the difference in potential energy that counts.
That's an interesting point. Is it safe to assume the zero was at the bottom you think?
 
Meeeessttteeehh said:
Thanks! No to the air resistance, I forgot about that... should I add it in? Or would it be "out of place" as I was taking about energy and air resistance isn't a type of energy? As for the conversion, the numbers suggest that the majority is converted to kinetic, but I guess this requires more research... Who knew skiing was so complicated!
Air resistance is a type of friction that dissipates mechanical energy to heat. This is probably more significant than the production of sound.

The final kinetic energy equals what percent of the loss of gravitational potential energy?
 
Meeeessttteeehh said:
That's an interesting point. Is it safe to assume the zero was at the bottom you think?
Most problem authors specify the zero of potential energy if it makes a difference to the answer. In this case, I think is safe to assume that zero is at the bottom of the hill. I thought I should point this out to you for future reference. Sometimes it's easier to write the appropriate equations with respect to one reference frame than another.
 
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