Energy - Ski sliding down a ramp

AI Thread Summary
A ski slides down a 28° incline that is 65m long, with a coefficient of friction of 0.090. The initial approach used energy relations to find the ski's speed at the base, but the calculation was incorrect due to misapplying the angle and friction force. The correct method involves equating potential energy to kinetic energy and the work done by friction, clarifying that energy dissipated equals the work done by friction. The discussion highlights the importance of using the correct angle and understanding the relationship between energy and force in calculations. The user plans to recalculate in the morning after addressing these errors.
Ritzycat
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Homework Statement


A ski starts from rest and slides down a 28 incline 65m long. If the coefficient of friction is 0.090, what is the ski's speed at the base of the incline? If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level?

Homework Equations


KE = (1/2)mv2
PEg = mgh

Note that I want to solve this problem using energy relations, not kinematics.

The Attempt at a Solution


Taking Point A to be the top of the ramp, point B to be right before the bottom.

PEg = KE + Ediss
mgh = (1/2)mv2 + Ff
mgh = (1/2)mv2 + μFN
mgh = (1/2)mv2 + μmg(sin 30)
gh = (1/2)v2 + μg(sin 30)
(9.8m/s2)(30.5157m) = (1/2)(v2) + (0.090)(9.8m/s2)(sin 30)

v = 24.4 m/s.

That answer was incorrect. I have not yet attempted the second question since I don't have a correct answer for the first one.
 
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Ritzycat said:

The Attempt at a Solution


Taking Point A to be the top of the ramp, point B to be right before the bottom.

PEg = KE + Ediss
mgh = (1/2)mv2 + Ff
mgh = (1/2)mv2 + μFN
mgh = (1/2)mv2 + μmg(sin 30)
gh = (1/2)v2 + μg(sin 30)
(9.8m/s2)(30.5157m) = (1/2)(v2) + (0.090)(9.8m/s2)(sin 30)

v = 24.4 m/s.

That answer was incorrect. I have not yet attempted the second question since I don't have a correct answer for the first one.

First thing, The angle of incline is 28 ,not 30. Secondly work done by friction is not μN.
 
Last edited:
Ritzycat said:

The Attempt at a Solution


Taking Point A to be the top of the ramp, point B to be right before the bottom.

PEg = KE + Ediss
mgh = (1/2)mv2 + Ff

The dissipated energy is not equal to the force of friction.
 
Tanya Sharma said:
First , The angle of incline is 28 ,not 30. Secondly work done by friction is not μFN.

Thanks for the reminder - although when I plugged in the new value, the answer changed very minimally...

ehild said:
The dissipated energy is not equal to the force of friction.

What should I set E diss equal to then??
 
Energy dissipated is equal to the work done by friction . How do you calculate the work done by a force ?
 
Ritzycat said:
Thanks for the reminder - although when I plugged in the new value, the answer changed very minimally...
What should I set E diss equal to then??
Do you think that energy is the same as force?
 
Tanya Sharma said:
Energy dissipated is equal to the work done by friction . How do you calculate the work done by a force ?

Ff * d = W

I think that will yield me the right answer. I would use the length of the incline, since that is the direction Friction was acting in. Then I'll plug it in. I'm right about to go to bed, so I'll do that in the morning.

ehild said:
Do you think that energy is the same as force?

No I forgot to multiply it by Distance....! i get it now Silly Ritzycat
 
Such things do happen :)
 
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