Solving for Cart Work, Speed & Friction - 10kg Cart

[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=??

 

[Doc Al]

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.

What's the definition of work?

 

[needhelp4rmu]

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

 

[Doc Al]

Work is force*distance

Good.

then i did 19*10=190 it doesn't work

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.

 

[needhelp4rmu]

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

 

[Doc Al]

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

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