Why Does the Toy Truck Stop at 3.2m Instead of 3.5m?

[Alex_]

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?

 

[dextercioby]

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.

 

[Alex_]

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]

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]

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.

 

[Alex_]

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]

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

Check your calculations again.

Daniel.

 

[Doc Al]

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.

 

[Alex_]

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.

 

[Alex_]

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

Check your calculations again.

Daniel.

where did you get the 180 over 57 from?

 

[Doc Al]

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 = ?