A single conservative force F = (7.0x - 11)i N, where x is in meters, acts on a particle moving along an x axis. The potential energy U associated with this force is assigned a value of 29 J at x = 0. (a) What is the maximum positive potential energy? At what (b) negative value and (c) positive value of x is the potential energy equal to zero?
(a) To find the maximum positive potential energy, we need to locate the minimum point of the potential energy function U(x). This occurs when the force F(x) is zero.
Given that the force F(x) is conservative, we know that the force can be derived from the potential energy function U(x) by taking the negative derivative:
F(x) = -dU(x)/dx
Substituting the given force F(x) into this equation:
7.0x - 11 = -dU(x)/dx
To find the potential energy function U(x), we integrate the force F(x) with respect to x:
U(x) = ∫(7.0x - 11)dx
U(x) = (7.0/2)x^2 - 11x + C
where C is the constant of integration.
Given that U = 29 J at x = 0, we can find the value of C:
29 = (7.0/2)(0)^2 - 11(0) + C
29 = C
Therefore, the potential energy function U(x) is:
U(x) = (7.0/2)x^2 - 11x + 29
To find the minimum point, we differentiate U(x) with respect to x and set the derivative equal to zero:
dU(x)/dx = 0
dU(x)/dx = (7.0/2)(2x) - 11 = 0
7.0x - 11 = 0
Solving this equation, we find:
x = 11/7
Substituting this value of x back into the potential energy function U(x), we find the maximum positive potential energy:
U(11/7) = (7.0/2)(11/7)^2 - 11(11/7) + 29 = 27 J
Therefore, the maximum positive potential energy is 27 J.
(b) To find the negative value of x where the potential energy is zero, we set U(x) equal to zero and solve for x:
(7.0/2)x^2 - 11x + 29 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 7.0/2, b = -11, c = 29
x = (-(-11) ± √((-11)^2 - 4(7.0/2)(29))) / (2(7.0/2))
x = (11 ± √(121 - 406)) / 7
x = (11 ± √(-285)) / 7
Since the discriminant is negative, we have no real solutions for x. Therefore, there are no negative values of x where the potential energy is zero.
(c) To find the positive value of x where the potential energy is zero, we set U(x) equal to zero and solve for x:
(7.0/2)x^2 - 11x + 29 = 0
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
a = 7.0/2, b = -11, c = 29
x = (-(-11) ± √((-11)^2 - 4(7.0/2)(29))) / (2(7.0/2))
x = (11 ± √(121 - 406)) / 7
x = (11 ± √(-285)) / 7
Again, since the discriminant is negative, we have no real solutions for x. Therefore, there are no positive values of x where the potential energy is zero.
To find the maximum positive potential energy, we need to determine the maximum positive value of the force F.
The force F is given by F = (7.0x - 11)i N, where x is in meters.
To find the maximum positive value of F, we need to find the value of x that maximizes the expression (7.0x - 11).
Since the coefficient of x is positive, the function (7.0x - 11) increases as x increases. Therefore, to find the maximum positive value of F, we need to find the maximum positive value of x.
To do this, we set the derivative of (7.0x - 11) equal to zero and solve for x:
d(7.0x - 11)/dx = 0
7.0 = 0
x = 11/7
So, the maximum positive value of x is x = 11/7 meters.
To find the maximum positive potential energy, we substitute this value of x into the equation for potential energy:
U = 29 J
Therefore, the maximum positive potential energy is 29 J.
To find the negative value of x at which the potential energy is zero, we need to solve the equation U = 0 for x:
29 = 0
x = -11/7 meters
Therefore, at x = -11/7 meters, the potential energy is zero.
To find the positive value of x at which the potential energy is zero, we again need to solve the equation U = 0 for x:
29 = 0
x = 11/7 meters
Therefore, at x = 11/7 meters, the potential energy is zero.
So, in summary:
(a) The maximum positive potential energy is 29 J.
(b) The negative value of x at which the potential energy is zero is x = -11/7 meters.
(c) The positive value of x at which the potential energy is zero is x = 11/7 meters.