- #1

- 1,197

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PE=Fd

0.5kx^2+3kx=Fd

0.5kx+3k=F

F=3k

The answer is -3k, why?

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- Thread starter UrbanXrisis
- Start date

- #1

- 1,197

- 1

PE=Fd

0.5kx^2+3kx=Fd

0.5kx+3k=F

F=3k

The answer is -3k, why?

- #2

- 13

- 0

Recall Work = -(delta)PE

Fd = -(0.5kx^2 + 3kx)

F = -0.5kx - 3k

F = -3k , x = 0

Fd = -(0.5kx^2 + 3kx)

F = -0.5kx - 3k

F = -3k , x = 0

- #3

- 1,197

- 1

what is work for KE?

W=.5mv^2

or

W=-.5mv^2

W=.5mv^2

or

W=-.5mv^2

- #4

Doc Al

Mentor

- 45,140

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The relationship between the potential function and the force is this: [itex]F = - dU/dx[/itex].

- #5

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- 1

would the object's Kinetic energy be positive then?

- #6

Doc Al

Mentor

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Beats me. That depends on the total mechanical energy, which you have not specified.UrbanXrisis said:would the object's Kinetic energy be positive then?

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