A single conservative force = (6.0x - 12) 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 23 J at x = 0.
(a) Write an expression for U as a function of x, with U in joules and x in meters.
U(x) =
(b) What is the maximum positive potential energy?
J
(c) At what negative value of x is the potential energy equal to zero?
m
(d) At what positive value of x is the potential energy equal to zero?
m
(a) U(x) = 23 J + 6.0x^2 - 12x
(b) The maximum positive potential energy occurs at the point where the derivative of U(x) with respect to x is zero. Differentiating U(x) with respect to x, we get:
dU/dx = 12x - 12
Setting dU/dx = 0 and solving for x:
12x - 12 = 0
x = 1
Substituting x = 1 into U(x), we get the maximum positive potential energy:
U(1) = 23 J + 6.0(1)^2 - 12(1) = 17 J
Therefore, the maximum positive potential energy is 17 J.
(c) To find the negative value of x at which the potential energy is zero, we need to set U(x) = 0 and solve for x:
23 + 6.0x^2 - 12x = 0
Using the quadratic formula, we find:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 6.0, b = -12, and c = 23. Plugging these values into the quadratic formula, we find that no real solutions exist. Therefore, there is no negative value of x at which the potential energy is zero.
(d) To find the positive value of x at which the potential energy is zero, we again set U(x) = 0 and solve for x:
23 + 6.0x^2 - 12x = 0
Using the quadratic formula, we find:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 6.0, b = -12, and c = 23. Plugging these values into the quadratic formula, we find no real solutions once again. Therefore, there is no positive value of x at which the potential energy is zero. It seems like this conservative force just doesn't like zero potential energy! Maybe it's trying to take a zero day or something.
(a) To find U(x), we need to integrate the given conservative force with respect to x.
Integrating the force (6.0x - 12) N, we get the potential energy expression:
U(x) = ∫(6.0x - 12) dx
Using the power rule of integration, we have:
U(x) = 3.0x^2 - 12x + C
Now, to find the constant C, we are given that U = 23 J at x = 0. Substituting these values into the expression:
23 = 3(0)^2 - 12(0) + C
C = 23
Therefore, the expression for U(x) is:
U(x) = 3.0x^2 - 12x + 23 J
(b) The maximum positive potential energy can be found by finding the maximum point of the quadratic function 3.0x^2 - 12x + 23.
To find the maximum, we can use the formula x = -b/2a, where a = 3.0 and b = -12.
x = -(-12) / (2 * 3.0)
x = 2
Substituting this value of x into U(x):
U(2) = 3.0(2)^2 - 12(2) + 23
U(2) = 12 - 24 + 23
U(2) = 11 J
Therefore, the maximum positive potential energy is 11 J.
(c) To find the negative value of x where the potential energy equals zero, we set U(x) = 0 and solve for x:
0 = 3.0x^2 - 12x + 23
This equation does not have real solutions, meaning the potential energy never equals zero for negative values of x in this case.
(d) To find the positive value of x where the potential energy equals zero, we set U(x) = 0 and solve for x:
0 = 3.0x^2 - 12x + 23
Using the quadratic formula, we get:
x = (-(-12) ± √((-12)^2 - 4 * 3.0 * 23)) / (2 * 3.0)
x = (12 ± √(144 - 276)) / 6.0
x = (12 ± √(-132)) / 6.0
Since the discriminant (√(-132)) is negative, this equation does not have real solutions. Hence, the potential energy never equals zero for positive values of x in this case.
(a) To express U as a function of x, we need to integrate the given conservative force. The potential energy function is given by the negative integral of the force function with respect to x:
U(x) = - ∫ (6.0x - 12) dx
Integrating the force function, we get:
U(x) = - (6.0 ∫ x dx - 12 ∫ dx)
Integrating each term:
U(x) = - (6.0 * (1/2)x^2 - 12x) + C
Simplifying and applying the given value U = 23 J at x = 0:
U(x) = - 3.0x^2 + 12x + 23 J
(b) To find the maximum positive potential energy, we can look for the vertex of the U(x) function, as this will give us the peak value. The vertex of a quadratic function of the form U(x) = ax^2 + bx + c is given by x = -b/2a.
In this case, a = -3.0 and b = 12. Plugging these values into the formula, we get:
x = -(12) / (2 * (-3.0)) = -12 / -6 = 2 meters
Using the value of x in the potential energy function:
U(2) = - 3.0(2)^2 + 12(2) + 23 J
U(2) = - 12 + 24 + 23 J
U(2) = 35 J
So the maximum positive potential energy is 35 J.
(c) To find the negative value of x at which the potential energy is equal to zero, we can set U(x) = 0 and solve for x.
Setting U(x) = 0:
0 = - 3.0x^2 + 12x + 23
We can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values from the equation:
x = (-(12) ± √((12)^2 - 4(-3.0)(23))) / (2 * (-3.0))
Simplifying under the square root:
x = (-(12) ± √(144 + 276)) / (2 * (-3.0))
x = (-(12) ± √(420)) / (-6)
x = (-(12) ± 2√(105)) / (-6)
x = (2 ± √(105)) / 3
Since we are looking for a negative value of x, we take the negative sign:
x = (2 - √(105)) / 3
Therefore, the negative value of x at which the potential energy is equal to zero is (2 - √(105)) / 3 meters.
(d) Similarly, to find the positive value of x at which the potential energy is equal to zero, we set U(x) = 0:
0 = - 3.0x^2 + 12x + 23
Using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
x = (-(12) ± √((12)^2 - 4(-3.0)(23))) / (2 * (-3.0))
Simplifying under the square root:
x = (-(12) ± √(144 + 276)) / (2 * (-3.0))
x = (-(12) ± √(420)) / (-6)
x = (-(12) ± 2√(105)) / (-6)
x = (2 ± √(105)) / 3
Since we are looking for a positive value of x, we take the positive sign:
x = (2 + √(105)) / 3
Therefore, the positive value of x at which the potential energy is equal to zero is (2 + √(105)) / 3 meters.
ΔU = U(x) - U(0) = ∫(6x - 12)dx
Therefore, with U(0) = 23 J, we obtain by integrating
U = 23 + 12x - 3x ²
F = 0 when 6x - 12 = 0
Thus x = 2 m
Substituting into U = 23 + 12x² - 3x ,
U = 35 J
Using the quadratic equation find the negative and positive values of x for which U = 0
3x ² - 12x-23 =0
x=(12±sqrt(144+4•3•23)}/2•6
x = -1.42 m
x = 5.42 m