The problem and solutions: http://i.imgur.com/hxlfbmZ.png
I was able to solve the first two questions, but I guessed part C.
For the first one, I thought of the two blocks as one unit (since there's no friction or air resistance). If they're both falling with the same acceleration, there shouldn't really be any tension in the rope. So the total work should be:
[itex]W_{tot}=\int^{0.75}_{0}(32x)dx[/itex] J
That's for the whole system. For parts A and B I did the above, but for one block a time.
Part C on the other hand... I can't seem to wrap my head around it. In my opinion, since we have friction, the only force felt by the 20 N block should be the 12 N in the direction of movement, and the friction [itex]20*µ_{k}=6.5[/itex] N in the opposite direction. I then want the total work in this case to be:
[itex]W_{tot}=\int^{0.75}_{0}(5.5x)dx[/itex] J
What am I doing wrong?!
I was able to solve the first two questions, but I guessed part C.
For the first one, I thought of the two blocks as one unit (since there's no friction or air resistance). If they're both falling with the same acceleration, there shouldn't really be any tension in the rope. So the total work should be:
[itex]W_{tot}=\int^{0.75}_{0}(32x)dx[/itex] J
That's for the whole system. For parts A and B I did the above, but for one block a time.
Part C on the other hand... I can't seem to wrap my head around it. In my opinion, since we have friction, the only force felt by the 20 N block should be the 12 N in the direction of movement, and the friction [itex]20*µ_{k}=6.5[/itex] N in the opposite direction. I then want the total work in this case to be:
[itex]W_{tot}=\int^{0.75}_{0}(5.5x)dx[/itex] J
What am I doing wrong?!
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