Conservative force and potential energy

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  • #1
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Problem:

A conservative force F is acting on a particle that moves vertically. F can be expressed as (3y-6) j head N, y is in m and j head is a unit vector along vertical direction.

a) Calculate the potential energy associated with F, with the potential energy set to zero at y=0.

b) At what values of y will the potential energy be maximum?

Relevant equations:

U(y) = - integral of F(y)

Attempt at the solution:

a) When F = 0, y = 2

U = - [integrate 0 to 2] (3y-6) = 6J

b) PE is maximum when F = 0. From (a), when F = 0, y = 2.


I am not familiar with this kind of questions. Could someone please tell me if it's correct?
 

Answers and Replies

  • #2
rude man
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In (a) you have not expressed U as a function of y.
In (b) you solved for max. U in a clever way, realizing that U will start to go down when F goes positive.
 
  • #3
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In (a) you have not expressed U as a function of y.

U = - integrate [0 to 2] F(y) = - integrate [0 to 2] (3y-6) = (3/2)y^2 - 6y.

Substituting the limits I get 6J. Is that correct?
 
  • #4
rude man
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Yes. More convetionally, y=2 for max. U is not known, so you would go

U = - ∫(3y - 6)dy = -3y2/2 + 6y + constant
But U = 0 when y = 0 so
0 = 0 + constant
constant = 0
Then set dU/dy = 0:
6 - (3/2)2y = 0
y = 2
You can check that U is a max, not a min, by computing d2U/dy2

with y = 2 to show that the second derivative is negative, meaning U is a max.
 

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