- #1
DeldotB
- 117
- 7
Homework Statement
Say you push a sled of mass [itex] m [/itex] up a hill that is angled upwards at a certain angle [itex] \theta [/itex] 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 [itex] \mu [/itex] added to [itex] mg sin \theta [/itex].
So: [itex] F_{push}= \mu + mgsin \theta=.30mg+mg sin \theta= mg(.30+sin \theta) [/itex].
If I push the sled down the hill with force [itex]F_{push} [/itex] 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:
[itex] \mu-F_{push}-mgsin \theta = ma [/itex] which means [itex] ma=.30mg-mg(.30+sin \theta ) -mg sin \theta [/itex]
So, the sled accelerates down the hill with acceleration [itex] a=-2g sin \theta [/itex].
Is this correct?