Conservative force and potential energy

[coconut62]

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?

 

[rude man]

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.

 

[coconut62]

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?

 

[rude man]

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.

 

[HalfLostAndSearching]

Your attempt at the solution is partially correct. Here is a more detailed explanation:

a) The potential energy associated with a conservative force is defined as the negative of the work done by the force in moving the particle from a reference point to a given position. In this case, the reference point is chosen to be y=0, so the potential energy at any point y is given by:

U(y) = -∫F(y)dy

Substituting the given force F = (3y-6)j, we get:

U(y) = -∫(3y-6)jdy

Integrating with respect to y, we get:

U(y) = -[3y^2/2 - 6y] + C

Since the potential energy is set to zero at y=0, we can find the value of the constant C by substituting y=0 in the above equation:

0 = -[3(0)^2/2 - 6(0)] + C

C = 0

Therefore, the potential energy associated with the given force F is given by:

U(y) = -[3y^2/2 - 6y]

b) To find the maximum potential energy, we can take the derivative of U(y) with respect to y and set it equal to zero:

dU(y)/dy = -3y + 6 = 0

Solving for y, we get y=2. This means that the potential energy will be maximum at y=2.

So, your answer for part (b) is correct, but your answer for part (a) is missing the constant C and the correct expression for potential energy. The correct answer is:

a) U(y) = -[3y^2/2 - 6y]

b) The potential energy will be maximum at y=2.