Grocery Cart Forces: Final Speed

[mortho]

Homework Statement

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?

Homework Equations

Wnet=deltaKE

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!

 

[rock.freak667]

The work done in moving the cart 11m is converted into 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

 

[mortho]

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

 

[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)

 

[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?

 

[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

 

[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 .