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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

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

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