Work done by Gravity

1. Jun 21, 2009

Amar.alchemy

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

Work and Kinetic energy theorem

3. 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.

2. Jun 21, 2009

Hootenanny

Staff Emeritus
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.

3. Jun 21, 2009

Amar.alchemy

so, if i assume that there is some friction present , with kinetic friction co-efficient MUk, then wat is the work done by gravity??

4. Jun 21, 2009

Hootenanny

Staff Emeritus
Well you'd first have to work out the work done by friction ...

5. Jun 21, 2009

Amar.alchemy

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...

6. Jun 21, 2009

Hootenanny

Staff Emeritus
Correct.
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?

7. Jun 21, 2009

Amar.alchemy

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: Jun 21, 2009
8. Jun 21, 2009

Hootenanny

Staff Emeritus
Correct.
Correct again, the change in potential energy of the body is always the product of the weight and the vertical distance raised.
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?

9. Jun 21, 2009

Amar.alchemy

Ya, in both cases the answer is "mgh"

Thank you very much Hootenanny :-)

10. Jun 21, 2009

Hootenanny

Staff Emeritus
A pleasure