# Friction Force Graph Problem

1. Oct 11, 2008

### jessedevin

1. The problem statement, all variables and given/known data

A 380 g cart is rolling along a straight track with velocity 0.850i m/s at x = 0. A student holds a magnet in front of the cart to temporarily pull forward on it, and then the cart runs into a dusting of sand that turns into a small pile. These effects are represented quantitatively by the graph of the x component of the net force on the cart as a function of position in the figure below.

(a) Will the cart roll all the way through the pile of sand? Explain how you can tell.

(b) If so, find the speed at which it exits at x = 7.00 cm. If not, what maximum x coordinate does it reach?
2. Relevant equations

W= (F)(d)cos(theta)
W= $$\int$$ F dx
KE= 1/2(m)(v)2

3. The attempt at a solution
I no that it does roll all the way through the pile of sand, but I am not sure why. Can someone please explain why.

To do the second part, what I did was use the kinetic energy formula above, and set the kinetic energy to the work involved. I found the amount of work done by finding the area under the curve of the graph, and adding each of the areas. For the work, i got:
W= $$\int$$ F dx
W= (2)(1) +1/2(4)(3=)= 8 Ncm --> .08 Nm
Then I set the kinetic energy equal to work
W=KE
.08= .5(.380kg)v2
v= .65 m/s

I know this is wrong, so can someone help me please!!!

Last edited: Oct 11, 2008
2. Oct 11, 2008

### LowlyPion

What is the initial KE of the cart at x=0?
Then what does each block of area on the graph represent?
Looks to me like each square represents 1*(.01)N-m = .01N-m

As the cart moves then area above the line is added to the Initial KE, area below is subtracted.

The velocity at any point then is given by converting the KE to mv2/2

3. Oct 11, 2008

### jessedevin

Well I first have to ask does the work equal the kinetic energy of the cart?

The initial KE f the cart is:
KE= .5mv2= .5(.38)(.850)2= .137 J
Okay, so now if I use this KE and add it with the KE i got from the graph, which is -.04, I get .097 J.
Then I use the KE formula again, so
.097=.5(.380)v2
v=.715 m/s
So thats right, i think, but can you explain to me why the cart roll all the way through the pile of sand?

4. Oct 11, 2008

### LowlyPion

If it gets to the other side and still has kinetic energy left over what else is it going to do?

5. Oct 11, 2008

### jessedevin

lol thanks...