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

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A 10 kg cart is subjected to varying forces over a distance of 10 meters, with discussions focusing on calculating net work done, final speed, and the coefficient of friction. The total work done is derived from summing the work for each segment, with the first segment yielding 84 Joules. Participants emphasize the importance of treating each segment separately, especially noting that the third segment involves a deceleration force. The conversation also addresses how to determine the speed of the cart after moving specific distances and the friction coefficient needed to bring the cart to a halt. Accurate calculations and understanding of work and forces are crucial for solving the problems presented.
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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=??
 
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needhelp4rmu said:
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?
 
Work is force*distance
then i did 19*10=190 it doesn't work
 
needhelp4rmu said:
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.
 
i tried it before. 84+8+3=95 it still didn't work..:(
 
needhelp4rmu said:
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?
 

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