A single conservative force Fx = (6.0x - 12) (x is in m) acts on a particle moving along the x axis. The potential energy associated with this force is assigned a value of + 20 J at x = 0. What is the potential energy at x = 3.0 m?
Not sure what this is all about. At x=0 Fx = 0 as well. No arbitrary assigned value.
To find the potential energy at x = 3.0 m, we need to calculate the work done by the conservative force between x = 0 and x = 3.0.
The potential energy associated with a conservative force can be calculated using the formula:
∆PE = -W
Where ∆PE is the change in potential energy and W is the work done by the conservative force.
To calculate the work done, we need to integrate the force over the displacement interval. The formula for calculating work done by a force is:
W = ∫F dx
In this case, the force Fx is given as 6.0x - 12.
So the equation for the work done becomes:
W = ∫(6.0x - 12) dx
To find the boundaries of the integral, we evaluate the force at the initial and final positions:
W = ∫(6.0x - 12) dx from x = 0 to x = 3.0
Now, let's integrate the force:
W = ∫(6.0x - 12) dx
= ∫6.0x dx - ∫12 dx
= 3.0x^2 - 12x (integrating term by term)
To calculate the change in potential energy (∆PE), we substitute the limits of integration:
∆PE = -W = -[3.0x^2 - 12x] from x = 0 to x = 3.0
Now, let's substitute the values of x into the expression:
∆PE = -[3.0(3.0)^2 - 12(3.0)] - [3.0(0)^2 - 12(0)]
Simplifying the expression:
∆PE = -[27 - 36]
∆PE = -[-9]
∆PE = 9 J
Therefore, at x = 3.0 m, the potential energy associated with the force is 9 J.