force vs kinematics

[Sonyfan08]

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If a skiier falls and begins sliding down the hill, and her speed at the bottom of the hill depends on: speed at the time of the fall, length of the hill and steepness of the hill. Why is weight not included? Wouldn't the forces acting on her be friction + gravitation (hence weight)?

 

[ruchir]

Are you sure there is friction? Skier on ice sounds like a classic frictionless problem 🙂

Also if you have the height of at top of the hill then the potential energy would convert into kinetic therefore the mass would cancel.

mgh = (mv^2)/2

m cancels, so you don't need it, assuming there is no friction.

 

[tmh]

I think you're right and just haven't carried it far enough.

If the person is going down at an angle there are two forces: friction and gravity.

The force of gravity is mgsinX (X being the angle)
The force of friction is u multiplied by the normal force, which is mgcosX. If you draw out the triangle you can see it.

Therefore the net forces are

mgsinX - umgcosX = ma

So mass cancels.

 

[deleted329605]

I know 1/2mv^2=mgh, and so masses cancel because they are on both sides. Any frictional forces you would have to subtract from initial potential energy before calculating speed. In terms of forces, you could figure out the speed assuming a constant acceleration (for this to be true, theta of incline has to be constant. You could construct a force diagram for the skier, with mgsin(theta) antiparallel to the frictional force (if it exists) and you'd get an constance acceloration setting it equal to ma. Then you could find the length and figure out what the final acceleration is using kinematic equations.

Remember that really, and the end of the day all of these things are related and can be derived from one another. like the change in KE is equal to the amount of work done.

Hope this helps!

 

[MT Headed]

If a person weighed more (i.e. more force) they would also have more mass.

F=ma

The acceleration would be the same no matter how much they weighed.

The equation for your problem is
(vf)^2 = (vi)^2 + 2ax
The factors you mentioned above affect (vi), x, and a (= g sin theta), respectively. But weight does not affect a.

 

[JRGhatt]

Would it be possible to solve this using PE = KE? If so, please show how.

Thanks!!

 

[muhali3]

Would it be possible to solve this using PE = KE? If so, please show how.

Thanks!!

You already did it.

What is PE? What is KE?

Set them equal like you did and solve for v.

(this assumes no friction though)

 

[JRGhatt]

gotcha. I did that but i guess i didnt make the connection between sqrt(2gh) and length and steepness of the hill.

thanks

 

[Majik]

Would it be possible to solve this using PE = KE? If so, please show how.

Thanks!!

Yes. Here's how:

Work Done by Friction = deltaME (where ME = Mechanical Energy); Rearranging this, we can solve for the final velocity:

(W. Friction) + PEi + KEi = PEf + KEf. At ground Level, PEf = 0J, therefore this simplifies down to:

(W. Friction) + PEi + KEi = KEf
mgcos(theta)mk + mgh + 1/2mvi^2 = 1/2mvf^2 (divide m out)
gcos(theta)mk + gh + 1/2vi^2 = 1/2vf^2

Rearranging, you can solve for final velocity. This question however asks for what variables alter the final velocity. We can see from the equation that:

- Final Velocity is independent of mass.
- 'h' = height from ground level, ie. the "length of the hill"
- theta = steepness of hill
- initial velocity also affects the final velocity

All these variables are important in calculating the final velocity, and therefore changing them could alter the magnitude of the final velocity.

 

[costales]

Another way of thinking: Gravity is a conservative force, whereas friction is a nonconservative force. So the work done by friction is just the difference in the total energy (gravitational potential + kinetic) at the two different elevations.