PHYC380
Classical Mechanics
Name _______________
Date ________________
Exercise 12: Work and Energy

1. Hooke’s Law for a spring of spring constant k is F   kx iˆ .
a. Show that it meets the two requirements for a conservative force.
a. The force depends only on the variable of position (x).

b.   F  0
i

x
−kx
j

y
0
k

z
0
b. Calculate its potential energy function U(x). Assume x = 0 is the
reference point, where U = 0.
x
1
U ( x)    ( kx) dx = kx 2
0
2
2. The figure to the right shows a plot of potential
energy U vs. position x of a 0.90-kg particle that can
travel only along the x axis. (Nonconservative forces are
not involved).
Three values are UA = 15.0 J, UB = 35.0 J,, and UC =
45.0 J. The particle is released at x = 4.5 m with an
initial speed of 7.0 m/s headed in the negative x
direction.
E = T + U = 1/2(0.9)(49) + 15 = 37.05 J
(a) If the particle can reach x = 1.0 m what is its speed there and if it cannot, where is its
turning point?
Can reach it. At x = 1.0 m, T = 37.05 − 35 = 2.05 J, v = 2.13 m/s − towards the left
What are the (b) magnitude and (c) direction of the force on the particle as it begins to
move to the left of x = 4.0 m? F = −d/dx U(x) = −(−20/2) = +10 N − towards the right
Suppose instead, the particle is headed in the positive x direction when it is released at x =
4.5 m at a speed 7.0 m/s.
E = T + U = 1/2(0.9)(49) + 15 = 37.05 J
(d) If the particle can reach x = 7.0 m what is its speed there, and if it cannot what is its
turning point?
Cannot reach it. The equation for U(x) at the point where T = 0 is y  15  30( x  5)
Set U(x) equal to  37.05 and solve for x.
Turning point is x = 5.74 m
What are the (e) magnitude and( f) direction of the force on the particle as it begins to
move to the right of x = 5.0 m?
F = −d/dx U(x) = −(30/1) = −30 N − towards the left