MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
Problem Solving Session 5 Conservation of Energy and Energy
Diagrams Solutions
IC_W08D1-1 Group Problem Spring-Loop-the-Loop Solution
A small block of mass m is pushed against a spring with spring constant k and held in
place with a catch. The spring compresses an unknown distance x . When the catch is
removed, the block leaves the spring and slides along a frictionless circular loop of radius
r . When the block reaches the top of the loop, the force of the loop on the block (the
normal force) is equal to twice the gravitational force on the mass. How far was the
spring initially compressed?
Solution:
We will use conservation of energy to find the kinetic energy of the block at the top of
the loop. We will then use Newton’s Second Law, to derive the equation of motion for
the block when it is at the top of the loop. Specifically, we will find the speed vtop in
terms of the gravitational constant g and the loop radius r . We will then combine these
results to find how far the spring was initially compressed.
Choice of Zero for Potential Energy: choose the gravitational potential energy to be zero
at the bottom of loop.
Initial Energy: Choose for the initial state the instant before the catch is released. The
initial kinetic energy is K 0  0 . The initial potential energy is non-zero, U0  (1 / 2)k x 2 .
The initial mechanical energy is then
E0  K 0  U0  (1 / 2)k x 2 .
(1)
Final Energy: Choose for the final state the instant the block is at the top of the loop. The
1
2
final kinetic energy is K f  m vtop
; the mass is in motion with speed vtop . The final
2
potential energy is non-zero, U f  (mg)(2R) . The final mechanical energy is then
E f  K f  U f  2mgR 
1
2
.
mvtop
2
(2)
Non-conservative Work: Since we are assuming the track is frictionless, there is no nonconservative work.
Change in Mechanical Energy: The change in mechanical energy is therefore zero,
0  Wnc  Emechanical  E f  E0 .
(3)
Mechanical energy is conserved, E f  E0 , or
1
1
2
2mgR  mvtop
 k x2 .
2
2
(4)
From Equation (4), the kinetic energy at the top of the loop is
1
1
2
mvtop
 k x 2  2mgR .
2
2
(5)
At the top of the loop, the forces on the block are the gravitational force of magnitude
mg and the normal force of magnitude N , both directed down. Newton’s Second Law
in the radial direction, which is the downward direction, is
mg  N  
2
m vtop
R
.
(6)
In this problem, we are given that when the block reaches the top of the loop, the force of
the loop on the block (the normal force, downward in this case) is equal to twice the
magnitude gravitational force on the block, N  2mg . The Second Law, Equation (6),
then becomes
2
m vtop
3 mg 
.
(7)
R
We can rewrite Equation (7) in terms of the kinetic energy as
3
1
2
mg R  m vtop
.
2
2
(8)
Combing Equations (5) and (8) yields
7
1
mg R  k x 2 .
2
2
(9)
Thus the initial displacement of the spring from equilibrium was
x
7mg R
.
k
(10)
IC_W08D1-2 Group Problem: Ball on String Solution
A ball of mass m is attached to point a by a string of length L . The ball is held out
horizontally as shown in the figure, then released from rest and allowed to fall. When the
string becomes vertical it impinges on a fixed peg at point b and the ball begins circular
motion about that point. What is the minimum distance D between points a and b
necessary to allow the ball to pass above b while still on a circular path? Note that the
alternative is for the string to go slack before the ball passes the line between a and b .
Solution:
From the geometry of the problem the radius of the orbit about the peg is equal to
R  ( L  D) .
At the top of its circular path about the peg the ball is a distance
L  2 R  L  2( L  D)  2 D  L below its starting point. Choose the starting point to be
the zero for the potential energy. Since there is no non-conservative work, mechanical
energy is constant throughout the motion and so at the top of the swing about the peg,
1
2
mvT 2  mg (2 D  L)  0 .
Now use F  ma at the top of the circular arc about the peg. Both the tension force of the
string acting on the ball (which we denote by T ) the gravitational force on the ball point
inward, and the acceleration points inward so the radial force equation becomes
mg  T  mvT 2 /( L  D) .
We are looking to find the minimum distance D between points a and b necessary to
allow the object to pass above b while still on a circular path. This occurs when the T
just goes to zero. Therefore
mg  mvT 2 /( L  D )
or rewriting this in terms of the kinetic energy we have that
mg ( L  D ) 1
 mvT 2 .
2
2
We can now combine this with our condition that the mechanical energy is constant
yielding
mg ( L  D)
 mg (2 D  L)  0 .
2
After a little rearranging of terms, the above equation becomes
3
5
mgL  mgD  0 .
2
2
We can solve this equation for the minimum distance D :
D
3
L.
5
IC_W08D1-3 Group Problem: Potential Energy Diagram
A body of mass m is moving along the x-axis. Its potential energy is given by the
function
U (x)  b(x 2  a 2 )2
(11)
where b  2 J  m-4 and a  1m .
Finding the x-component of the Force from the Potential Energy:
Recall the definition of potential energy
x
U (x)  U (x0 )    Fx ( x ) dx
x0
(12)
The first fundamental theorem of calculus states that
U (x)  U (x0 ) 

x
x0
dU ( x )
dx .
dx
(13)
Comparing Equation (12) with Equation (13) shows that the x -component of the force is
the negative derivative (with respect to position) of the potential energy,
Fx (x) 
dU (x)
.
dx
(14)
Question 1: What is the x -component of the force associated with the potential energy
given by Equation (11)?
Answer:
To find the x -component of the force from the potential we use
Fx (x) 
dU (x)
 4b(x 2  a 2 )x
dx
Question 2 Make a sketch of the x -component of the force Fx (x) vs. x .
Extrema Points:
The minimum and maximum points (the extrema) of the potential energy function occurs
at the point where the first derivative vanishes
dU ( x )
 0.
dx
(15)
Question 3: Find the minimum and maximum points of the potential energy.
Answer
The extrema points of the potential occur when
dU ( x )
0
dx
or equivalently when
0  Fx (x)  2b(x 2  a2 )2x .
There are three extrema at x  0 , x  a , and x  a .
Stable Equilibrium Points:
Suppose the potential energy function has positive curvature in the neighborhood of a
extremum at x0 . If the body is moved a small distance to the point x  x0 away from this
point x0 , the slope of the potential energy function is positive, dU ( x) dx  0 ; hence the
component of the force is negative since Fx   dU ( x) dx  0 . Thus the body
experiences a restoring force towards the extremum point of the potential. If the body is
moved a small distance to the point x  x0 then dU ( x) dx  0 ; the component of the
force is positive since Fx   dU (x) dx  0 . Thus the body again experiences a restoring
force towards the extremum point of the potential.
Question 4 Which of these extrema points correspond to a stable equilibrium point?
Answer:
The points x  a and x  a correspond to stable equilibrium points because if you
move slightly away from these points the force is restoring, i.e. the direction of the force
is always points towards the equilibrium point. The point x  0 is an unstable
equilibrium point. If you move slightly away from x  0 in either direction, the force
points away from x  0 . For example consider values of x  0 , from the graph of Fx (x)
vs. x , the sign of the x-component of the force is positive hence the force points in the
positive x-direction which is away from x  0 . If the object moves in the negative xdirection, then the x-component of the force is negative, which is again pointing away
from x  0 . So x  0 is an unstable equilibrium point. You can also use the potential
energy graph to determine that x  0 is an unstable equilibrium point.
towards the extrema of the potential energy. Such an extremum point is called a stable
equilibrium point.
Energy Diagram:
The mechanical energy at any point x is the sum of the kinetic energy K (x) and the
potential energy U (x)
E(x)  U (x)  K(x)
(16)
Both the kinetic energy and the potential energy are functions of the position of the body.
Assume that the mechanical energy is a constant E , then for all points x
E(x)  E
(17)
The energy is a constant of the motion and can be either a positive value or zero. When
the energy is zero, the body is at rest at the equilibrium positions, x  a or x  a .
A straight horizontal line on the plot of U (x) vs. x corresponds to a non-zero positive
value for the energy E . The kinetic energy at a point x1 is the difference between the
energy and the potential energy,
K (x1 )  E  U (x1 )
(18)
At the points, where E  U (x) , the kinetic energy is zero. Regions where the kinetic
energy is negative, are called the classically forbidden regions, which the body can never
reach if subject to the laws of classical mechanics. In quantum mechanics, there is a very
small probability that the body can be found in the classically forbidden regions.
Question 5: Describe what happens to the body if it has energy E  2 J and is located at
at x  0 .
Answer: At x  0 the body has a positive kinetic energy hence is moving in either the
positive or negative x -direction. As the object moves away from x  0 , the kinetic
energy increases, whereas if the object moves slightly away from x  a or x  a , the
kinetic energy decreases hence the speed decreases and the object slows down until it
eventually stops where K  0 . However the x -component of the force is non-zero at
those points (slope of U (x) vs x is non-zero), hence it reverses direction.
Question 6: Suppose the body starts with zero speed at x  4a . What is its speed at
x  0 and at x  a ?
Answer:
At x  4a , K (4a)  0 . Also
U (4a)  b(16a 2  a 2 )2  225ba 4  225(2 J  m-4 )(1m)4  450 J .
Therefore the value of the energy is
E  E (4a)  K (4a)  U (4a)  0  450 J=450 J .
At x  0
U (0)  b(a 2 )2  ba 4  (2 J  m-4 )(a  1m)4  2J .
So
K (0)  E  U (0)  450 J - 2 J = 448 J .
Therefore at x  0 the speed is
v(0) 
2 K (0)
896 J
.

m
m
Similarly at x  a , U (a)  0 . So
K (a)  E  U (a)  450 J  0 = 450 J .
Therefore at x  a , the speed is
v(a ) 
2 K ( a)
900 J
.

m
m