You have a level track. You push a cart with mass = 0.82[kg]. You measure the in

Q: You have a level track. You push a cart with mass = 0.82[kg]. You measure the in

We know that from the work energy theorem Work done = Chnage in kinetic energy =(1/2)m(v2-u2) =0.5*0.82*((0.458)2 -(0.61)2) =0.0665 =-66.584mJ

ID: 1367419 • Letter: Y

Question

You have a level track. You push a cart with mass = 0.82[kg].
You measure the initial velocity to be 0.61[m s-1].
2 seconds later, you measure the velocity to be 0.458[m s-1].
What is the work (reported in mJ) that friction did on the cart?
work = ? [mJ]

I came up with -38.0 mJ but it was wrong any help would be appreciated

these instructions also came with the question:
Use three sig. figs. or N/A if not enough information is given, such as change in position!
Remember, that if the system gains energy, then positive work is done;
otherwise the system loses energy work done on system < 0.

Explanation / Answer

We know that from the work energy theorem

Work done = Chnage in kinetic energy =(1/2)m(v2-u2) =0.5*0.82*((0.458)2 -(0.61)2) =0.0665 =-66.584mJ

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You have a level track. You push a cart with mass = 0.82[kg]. You measure the in

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