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I can find the work done in the 2nd case using W=fd= 39.2 J.

- Thread starter pinkyjoshi65
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- #1

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I can find the work done in the 2nd case using W=fd= 39.2 J.

- #2

Astronuc

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What is the weight of a 4.00 kg object? Remember weight is a force (due to gravity).

- #3

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oh yes, W=mgd= 4*9.8*1= 39.2J

So the work done is the same!

So the work done is the same!

- #4

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You raise a 2.00-kg pail of water 1.00 m and then lower it 1.00 m. What is the total work done?

W=mgd

W=2*9.8*1=19.6J

Thats the work done when the pail is lifted.

The work done when the pail is lowered is

W= mgcos(theta)d

and theta is 180

so that would make w= -19.6J

so total work will be zero. yes..?

- #5

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opps..the question is what is the total work done by gravity

- #6

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So if the initial and final positions as in this case are the same, the work will be the same. You don't need to calculate anything.

- #7

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- #8

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so the total work done will be 19.46J..?

- #9

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oh..i get it..so its -19.6J..thankz..!

- #10

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- #11

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- #12

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That isn't the right thing to say. Work is a scalar quantity given by the dot product(scalar product of two vectors) [tex] \vec{W}= \vec{F} . \vec{s} [/tex]. So it depends on two parameters(viz. the component of the force responsible for the work and the displacement in that direction) and not just one. You'd have to have a condition for both the parameters. What you wanted to say I guess is that knowing the magnitude of a constant force, F and a constant displacement, s ; you are merely changing the direction of the force so that only one of its component is responsible for the work. In that case, yes, you are right. But you have to take care about what you assume.

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