# Cart with Friction

1. Oct 8, 2008

### needhelp4rmu

A 10 kg cart is pulled by a force of 12 newtons for a distance of 7 meters, then by a force of 4 Newtons for 2 meters, then it is pushed (slowed down) by a force of 3 Newton for 1 meters. All this occurs on a horizontal surface.

A)Initially assuming the surface to be frictionless, what net work is done on the cart as it moved the 10 meters?
Work total?
i know work total = (.5mvfinall^2)-(.5mvinitial^2) I don't know what to plug in for velocity..im stuck here...cause i got to go to work...help please.

B)If the cart starts from rest, what is its speed after it has moved the 10 m?
Vf=?

Now assume a coefficient of friction µ between cart and surface during the entire distance.
c) For what value of μ does the cart just come to a halt after 10 m?

µ =??
D) For the value of µ computed in part (c), what is the speed of the cart after it has gone 4 m?
v=??

2. Oct 8, 2008

### Staff: Mentor

What's the definition of work?

3. Oct 8, 2008

### needhelp4rmu

Work is force*distance
then i did 19*10=190 it doesn't work

4. Oct 9, 2008

### Staff: Mentor

Good.
No. Do each segment separately and add them up to find the total work done. For example, the first segment has F = +12 and distance = +7, so the work done for that segment would be: (12)*(7) = +84 Joules. Keep going.

5. Oct 9, 2008

### needhelp4rmu

i tried it before. 84+8+3=95 it still didn't work..:(

6. Oct 9, 2008

### Staff: Mentor

The third segment is incorrect. What's different about it compared to the first two?