# Need Help With a Problem in Physics 30

I can't get this problem and wanted to see if anyone could help me out:

A toy truck has a velocity of 6.0 m/s when it begins to role freely up a ramp inclined at 30.0 degrees. They toy has a mass of 5.0 kg, and the frictional forces present are 4.0N. What distance does the truck travel before stopping?

I keep on getting 3.5m but the correct answer is 3.2m. Anyone know how to solve this?

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dextercioby
Homework Helper
How about using the theorem of variation of KE...?I don't know if you have,but i found this to be the most elegant way

Daniel.

Well I've never heard of that theorem before, but maybe I've used it though. What I tried to do to solve this problem is figure out total energy before (Et = Ek + Ep) and then solve for when Ek is 0 after the toy has rolled up the ramp. The thing that complicates this is there's friction involved(heat, F * d). So I'm not sure how I have to include friction into this problem in order to get the right answer.

Doc Al
Mentor
Set the initial energy (purely KE, if you measure PE from the bottom of the ramp) equal to the final energy (PE + the heat due to friction = PE + F*d):
$${KE}_i = {PE}_f + F*d$$

dextercioby
Homework Helper
For the record the theorem (proven in an elementary version by G.W.Leibniz) states that the variation of KE for a mechanical system is equal to the total work done by external forces acting on the system.In your case,the forces are friction & gravity and the mechanical system is composed only of one particle/body.

Daniel.

Doc Al said:
Set the initial energy (purely KE, if you measure PE from the bottom of the ramp) equal to the final energy (PE + the heat due to friction = PE + F*d):
$${KE}_i = {PE}_f + F*d$$
I tried to do that, but got 3.5 as my answer, so I was off by a little bit. I'll show you how I did this problem in deatil, maybe you'll be able to catch any mistakes?

Et_i = Ek_i
Ek_i = 0.5 * 5 * 6^2
Ek_i = 90

Et_f = Ep_f + (F * d)
90 = -5 * -9.81 * h + (4 * (sin30/h))
h = 1.75 (approx.)

sin30 = 1.75/x
x = 3.5 (approx.)

dextercioby
Homework Helper
The work done by friction is negative as well.Like the one performed by gravity.The lenght is approximately
$$x\sim \frac{180}{57}m\sim 3.2 m$$

Daniel.

Doc Al
Mentor
Alex_ said:
Et_f = Ep_f + (F * d)
90 = -5 * -9.81 * h + (4 * (sin30/h))
h = 1.75 (approx.)
Redo this calculation. x = h/sin30; you have it reversed.

Doc Al said:
Redo this calculation. x = h/sin30; you have it reversed.
I tried that and ended up getting 4.38, so that doesn't work either.

dextercioby said:
The work done by friction is negative as well.Like the one performed by gravity.The lenght is approximately
$$x\sim \frac{180}{57}m\sim 3.2 m$$

Daniel.
where did you get the 180 over 57 from?

Doc Al
Mentor
Alex_ said:
I tried that and ended up getting 4.38, so that doesn't work either.
Well, try it again. You are making an arithmetic mistake:
90 = (5)(9.8)h + (4)[2h] = 57h
=> h = ?