# Net work

Discussion in 'MCAT Study Question Q&A' started by texan2414, 09.30.14.

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1. ### texan2414 2+ Year Member

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Was doing a problem in TBR yesterday where in an elliptical orbit, the work done was non-zero.
Now I know that since the object's velocity increased as it approached the Sun, there must be some work done to change its energy but what if it had been a perfectly circular question? Does work done in that case is still non-zero with total NET work being done zero?

Would love a clarification on that-

Thanks,

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3. ### Cawolf 2+ Year Member

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Uniform circular motion as constrained by gravity or tension in a string involves no net work as you said.

Work is proportional to energy which is proportional to a change in the magnitude of the velocity in this case.

The magnitude in uniform circular motion is constant, only the direction is changing.

Another way to look at it is that W = F dot d (if you know the dot product) which is the same as saying W = Fdcos(theta).

The force is always perpendicular to the displacement in this type of motion and the cos (90) = 0 so no work is done.

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4. ### texan2414 2+ Year Member

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Thanks,