Sledding Down a Hill - Speed Calculation

[DeldotB]

Homework Statement

Say you push a sled of mass m up a hill that is angled upwards at a certain angle \theta at a constant velocity. The hill has snow on it offering a friction force that is equal to 30% of the sleds weight. If you pushed the sled down the hill with the same amount of force as you did pushing the sled up the hill, how fast does it go down the hill?

Homework Equations

Newtons Laws

The Attempt at a Solution

So, since the sled is moving up the hill at a constant velocity, the net force on the sled must be zero.
This also means that the force with which you push is equal to the friction force \mu added to mg sin \theta.

So: F_{push}= \mu + mgsin \theta=.30mg+mg sin \theta= mg(.30+sin \theta).

If I push the sled down the hill with force F_{push} then the sled will accelerate.

Since the frictional force now points in the opposite direction when I push the sled down the hill, I will have:

\mu-F_{push}-mgsin \theta = ma which means ma=.30mg-mg(.30+sin \theta ) -mg sin \theta

So, the sled accelerates down the hill with acceleration a=-2g sin \theta.

Is this correct?

 

[gneill]

Looks good. You might want to finish with an expression for the velocity with respect to time, as the problem did ask for how fast it goes.

 

[DeldotB]

Looks good. You might want to finish with an expression for the velocity with respect to time, as the problem did ask for how fast it goes.

Shall I integrate a?

 

[gneill]

You could. Although, since the acceleration is constant you might just use a standard kinematic formula.

 

[DeldotB]

Thanks for the help gneill

 

[haruspex]

$\mu-F_{push}-mgsin \theta = ma$

Yes, if you are taking the positive direction for a as up the slope (which is why you got a negative acceleration at the end).