Question about work, energy and the work energy theorem.

[Ethan_Tab]

Homework Statement

So a mass m is placed on a ramp at a height given by h_1. This portion of the trip is frictionless. Then, at the bottom of the ramp the mass encounters a rough patch of levelled ground with a µk given my µ. This strip has a length of size "d" meters. After traveling across this surface, the mass encounters yet again, another frictionless ramp. The question wants to know how far up the second ramp the ball will get. Call this variable h_2. There were no ø given for either of the two ramps.

Homework Equations

ΔE=W
GPE=mgh
KE=½mv^2
W+KE_1+GPE_1=KE_2+GPE_2+E_loss (due to friction)
KE_1=0
KE_2=0 (when it reaches it max height on the second hill)

The Attempt at a Solution

If someone could confirm this or explain why this is incorrect it would be very helpful.

Im thinking,
Since W=F*D and I know the only force acting against the ball is that of friction (negative work), can I equate that to ΔGPE?

So; -µmgd=mgh_2-mgh_1
-µd+h_1=h_2

But now that I think about it, would there also be work done by the x component of gravity when the ball is rolling down and up the first and second hills? If anyone could clarify this it would be much appreciated.

 

[AlephNumbers]

Your solution looks correct to me.

But now that I think about it, would there also be work done by the x component of gravity when the ball is rolling down and up the first and second hills?

Could you elaborate upon this? I am not entirely sure what you are confused about.

 

[Ethan_Tab]

Your solution looks correct to me.

Could you elaborate upon this? I am not entirely sure what you are confused about.

Can I just assume all work done is that which the mass encountered when it slid across the rough patch of length "d" or do I have to count for the work done by gravity when the mass was on both ramps?

 

[AlephNumbers]

-µmgd=mgh_2-mgh_1

It looks like you did account for it.

Gravity is a conservative force. The work done on the mass by the force of gravity is equal to the change in kinetic energy of the mass; W = ΔK

 

[Ethan_Tab]

It looks like you did account for it.

Gravity is a conservative force. The work done on the mass by the force of gravity is equal to the change in kinetic energy of the mass; W = ΔK

Oh! I see! Thanks for the help .