Final answer:
The equilibrium position of a particle with potential energy given by U(x) = ax - bx^2 is found by setting the derivative of U(x) equal to zero and solving for x, resulting in the equilibrium position x = a/(2b).
Step-by-step explanation:
To find the equilibrium position of a particle moving along the x-axis where its potential energy is given by U(x) = ax - bx^2, we must set the derivative of the potential energy with respect to x equal to zero, as the equilibrium position occurs where the force is zero, which corresponds to the potential energy being at a minimum or maximum.
Let's calculate the derivative of the potential energy function:
U'(x) = d/dx(ax - bx^2) = a - 2bx
Now, set this derivative equal to zero to find the equilibrium:
a - 2bx = 0
By solving for x, we get:
x = a/(2b)
This x is the equilibrium position of the particle.
To find the equilibrium position of a particle with potential energy given by U(x) = ax - bx^2, we need to determine the point where the force on the particle is zero. The force is given by F(x) = -dU/dx. So, we can find the equilibrium position by setting F(x) equal to zero and solving for x.
First, differentiate U(x) with respect to x to find dU/dx. Then, set dU/dx equal to zero and solve for x to find the equilibrium position.
For the given potential energy function U(x) = ax - bx^2, where a and b are constants, the equilibrium position can be found by differentiating U(x) with respect to x:
dU/dx = a - 2bx
Setting dU/dx equal to zero:
a - 2bx = 0
2bx = a
x = a/(2b)
So, the equilibrium position of the particle is x = a/(2b).