# Grocery cart forces

1. Dec 13, 2007

### mortho

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

A 105.0 N grocery cart is pushed 11.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all frictional forces are neglected and the cart starts from rest, what is the grocery cart's final speed?

2. Relevant equations

Wnet=deltaKE

3. The attempt at a solution

So i used that equation which ended up to mad=1/2mv2

(10.7)(9.81)(11)=1/2(10.7)(v2)
v= 14.7 m/s

But it ended up being wrong, i don't get it. and also i could cancel out the masses right? but it would still give me same answer so it wouldn't matter. Thanks for your help!

2. Dec 13, 2007

### rock.freak667

The work done in moving the cart 11m is converted in to k.e.
so that $Fd=\frac{1}{2}mv^2$
so F is the constant force exerted to move the cart 11m(d) and m is its mass(105/9.81) and re-arrange to find v^2

3. Dec 13, 2007

### mortho

wait isn't that wat i did though..Fd=1/2mv2 ?

4. Dec 13, 2007

### cryptoguy

In F*d, F refers to the Horizontal force (Fa) aka 40N. Now for 1/2*mv^2, you use the M of the cart (10.7)

5. Dec 13, 2007

### mortho

oh? my teacher was saying use force parallel, oohhh..but i guess that would be the same as Force applied. ok so i worked it out as (40)(11)=1/2(10.7)v2
my answer is 9.07 m/s correct?

6. Dec 13, 2007

### cryptoguy

Right, in Work problems, you consider the part of the force that's acting parallel to the distance it's acting on, in this case that force is 40 N. 9.07 looks right

7. Dec 13, 2007

### mortho

THANK YOU! hey could you help me out with a skier problem ,,,it's on the bulletin,,i think it's called skier coefficient or velocity..that one's soo hard and it's almost due .