he potential energy associated with a conservative force is given by the equation:
U(x) = - ∫F(x)dx + C
where F(x) is the force as a function of position, dx is an infinitesimal displacement, and C is an arbitrary constant of integration.
For the given force F = (4.0x - 11) N, the potential energy U(x) can be found by integrating:
U(x) = - ∫F(x)dx + C
U(x) = - ∫(4.0x - 11)dx + C
U(x) = -2.0x^2 + 11x + C
To find the value of the constant C, we use the given information that U = 27 J when x = 0:
U(0) = -2.0(0)^2 + 11(0) + C = 27 J
C = 27 J
So the potential energy is:
U(x) = -2.0x^2 + 11x + 27
(a) The maximum positive potential energy occurs at the vertex of the parabolic function -2.0x^2 + 11x + 27. The x-coordinate of the vertex can be found using the formula -b/(2a), where a = -2.0 and b = 11:
x = -b/(2a) = -11/(2(-2.0)) = 2.75 m
To find the maximum positive potential energy, we substitute this value of x into the potential energy equation:
U(2.75) = -2.0(2.75)^2 + 11(2.75) + 27 ≈ 38.375 J
So the maximum positive potential energy is approximately 38.375 J.
(b) The potential energy is zero when:
-2.0x^2 + 11x + 27 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = -2.0, b = 11, and c = 27.
Plugging in these values, we get:
x = (-11 ± sqrt(11^2 - 4(-2.0)(27))) / 2(-2.0)
x ≈ -2.23 or x ≈ 6.13
So the potential energy is zero at x ≈ -2.23 m and x ≈ 6.13 m.
(c) Since the force F is conservative, the potential energy is symmetric around the x-axis. Therefore, the potential energy is also zero at the same distances from x = 0, but on the opposite side of the x-axis. So the potential energy is zero at x ≈ 2.23 m and x ≈ -6.13 m.