# Homework Help: Cart with Friction

1. Oct 13, 2009

### r34racer01

A 10 kg cart is pulled by a force of 14 newtons for a distance of 6 meters, then by a force of 6 Newtons for 4 meters, then it is pushed (slowed down) by a force of 6 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 11 meters?
b) If the cart starts from rest, what is its speed after it has moved the 11 m?

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

W = ΔKE
W = Fd cosθ
Ff = μN
ΣF = ma
KE = (1/2)mv^2

So for a.) I got 101 joules by dumb luck. I thought I could do Wtot = (14*6)+(6*4)+(6*1) = 114, but that didn't work and I just entered 101 at random. Can someone tell me how to actually calculate the total work?
For b.) I knew Wtot = (1/2)mv^2, so 101=5v^2 => Vf= 4.49

c.) is where I'm really stuck. I thought I could do Wtot = Fd = μ*m*g*d => μ = 0.094
but that's wrong. Can someone tell me how to get this part please?

2. Oct 13, 2009

### tiny-tim

Hi r34racer01!
The third force is backwards

(but I still don't make it 101)

3. Oct 13, 2009

### r34racer01

Yeah apparently it is actually W = 102 for pt a.), weird right. Anyway I'm still having trouble w/ pt c.). It should be 102 = μmgd = 0.259 but its telling me that's wrong. Help anyone?

4. Oct 13, 2009

### r34racer01

Never mind I got it, I was plugging in the wrong numbers.