Solving Toboggan Homework: Work & Kinetic Energy

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The discussion revolves around calculating the vertical height a toboggan reaches on a frictionless hill after being propelled at a speed of 12 m/s. The key equation derived is y = v²/2g, indicating that the vertical height is independent of the slope's angle. When friction is introduced, the work done against gravity and friction must be considered, but the change in gravitational potential energy remains dependent solely on the vertical height raised. Clarifications emphasize that the work done by gravity is consistently mgh, regardless of the slope's angle. The conversation concludes with an affirmation of the principles of work and energy in this context.
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Homework Statement



At the base of a frictionless icy hill that rises at 25 degree above the horizontal, a toboggan has a speed of 12 m/s toward the hill. How high vertically above the base will it go before stopping?

Homework Equations



Work and Kinetic energy theorem

The Attempt at a Solution



W = −mgy, where y is the vertical height.
−mgy=−1/2 mv2 and
therefore y equals v2/2g

it means vertical height is independent of slope. Is this correct answer?? Bcoz if i assume that slope is zero then also it should reach the vertical height given by the above equation. Kindly clarify.
 
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Amar.alchemy said:

Homework Statement



At the base of a frictionless icy hill that rises at 25 degree above the horizontal, a toboggan has a speed of 12 m/s toward the hill. How high vertically above the base will it go before stopping?

Homework Equations



Work and Kinetic energy theorem

The Attempt at a Solution



W = −mgy, where y is the vertical height.
−mgy=−1/2 mv2 and
therefore y equals v2/2g

it means vertical height is independent of slope. Is this correct answer?? Bcoz if i assume that slope is zero then also it should reach the vertical height given by the above equation. Kindly clarify.
That is indeed the correct answer. The angle of the slope would be important if the slope was frictionless, however, since the slope is indeed frictionless the vertical height should be independent of the inclination. The distance parallel to the slope will depend on the inclination.

If the slope was zero, your equation wouldn't make much sense now would it because it would never leave the horizontal.
 
so, if i assume that there is some friction present , with kinetic friction co-efficient MUk, then wat is the work done by gravity??
 
Amar.alchemy said:
so, if i assume that there is some friction present , with kinetic friction co-efficient MUk, then wat is the work done by gravity??
Well you'd first have to work out the work done by friction ...
 
Work done by friction comes out to be -mukmgh / tan alpha

ok, so here comes the angle alpha, so what i am not understanding is why we are not considering alpha while calculating work done by gravity?? sorry if i am dragging this too much...
 
Amar.alchemy said:
Work done by friction comes out to be -mukmgh / tan alpha
Correct.
Amar.alchemy said:
ok, so here comes the angle alpha, so what i am not understanding is why we are not considering alpha while calculating work done by gravity?? sorry if i am dragging this too much...
Okay, so if the hill has friction, then the kinetic energy can be transformed into the work done against gravity and work done against the friction. However, for the case of a frictionless hill, the kinetic energy is only transferred into gravitational potential energy. What is the only way to increase a body's gravitational potential energy?
 
Then i have to raise its height against gravity... rite??

But still the work done by gravity in the "hill with friction is mgh"... rite??

what i mean is work done by gravity should be "mgh sin alpha"
 
Last edited:
Amar.alchemy said:
Then i have to raise its height against gravity... rite??
Correct.
Amar.alchemy said:
But still the work done by gravity in the "hill with friction is mgh"... rite??
Correct again, the change in potential energy of the body is always the product of the weight and the vertical distance raised.
Amar.alchemy said:
what i mean is work done by gravity should be "mgh sin alpha"
Why should that be the case?

Let's take an example. Can you answer the following questions:

(1) Suppose I have a mass m and I raise it vertically h meters. What is the change in gravitational potential energy of the mass?

(2) Suppose that once again I have a mass m and I raise it vertically h meters. Then I move it h meters horizontally. What is the change in gravitational potential energy of the mass?
 
Ya, in both cases the answer is "mgh"

Thank you very much Hootenanny :-)
 
  • #10
Amar.alchemy said:
Ya, in both cases the answer is "mgh"

Thank you very much Hootenanny :-)
A pleasure :smile:
 

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