MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
Problem Solving Session 4: Work and Kinetic Energy Solutions
Problem 1 Collision and Sliding on a Rough Surface
Block 1 of mass 3m is sliding along a frictionless horizontal table to the right with speed
v0 . Block 1 collides with block 2 of mass m that is moving to the left with speed v0 .
After the collision, the two blocks stick together and the blocks enter a rough surface at
x  0 with a coefficient of kinetic friction that increases with distance as k (x)  bx 2 for
0  x  d , where b is a positive constant. The blocks come to rest at x  d . The
downward gravitational acceleration has magnitude g . Determine an expression for the
initial speed v0 of the blocks in terms of b , d , m , and g as needed.
Solution: Because there are no external forces in the x -direction, momentum is constant
during the collision hence
3mv0  mv0  4mva .
So the speed immediately after the collision is
va  v0 / 2 .
The work done by the friction force on the blocks as they move together along the surface
is given by
xd
W 
 bx
2
x0
The kinetic energy after the collision is
4mg dx  (4 / 3)mgbd 3 .
K a  (1 / 2)4mva2  (1 / 2)4m(v02 / 4)  (1 / 2)mv02 .
The final kinetic energy is zero. So the work energy theorem, W  K f  K a , becomes
(4 / 3)mgbd 3  (1 / 2)mv02 .
Thus the speed of the blocks before the collision is
v0  (8 / 3)gbd 3 .
Problem 2 Sliding Along a Sphere
An object of mass m initially sits on top of a large sphere of radius R that is fixed to the
ground as shown in the figure. The object gets a small push and begins to slide along the
surface of the sphere. You may assume that the initial kinetic energy is negligible. There
is a friction force between the object and the surface that varies with the angle 
according to f  f0 sin  where f0  mg is a constant. Let g denote the magnitude of
acceleration due to gravity.
The object just loses contact with the surface of the sphere at an angle  f with respect to
the vertical when it has a speed v f .
a) What is the work done by the friction force on the object as the object moves
through the angle   0 to the angle    f ? Hint: the small displacement of the
r
object on the surface of the sphere is given by dr  Rd φ.
r
Solution: The friction force is f   f0 sin φ. The work done by the friction force is
   f
r r   0
W   f  d r    f0 sin   φ R d  φ   f0 R
f
f
i
  0

sin   d 
  0
   f
W  f0 Rcos    0  f0 R(cos f  1)
f
b) What is the work done by the gravitational force on the object as the object moves
through the angle   0 to the angle    f ?
r
Solution: The gravitational force is mg  mg sin  φ  mg cos rφ. The work done by the
gravitational force is
f
  0
i
  0
r r
W   mg  d r 
g

(mg sin   φ  mg cos  rφ)  R d  φ  mg
   f

sin   d 
  0
   f
W g  mgRcos    0  mgR(cos f  1)
c) Based on your results from parts a) and b), use the work-energy theorem to
determine an expression for the kinetic energy of the object just before the object
leaves the surface of the sphere in terms of  f , m , g , R , f 0 and v f as needed.
Solution: The sum of the work is
W  W f  W g  mgR(cos f  1)  f0 R(cos f  1)  R(1  cos f )(mg  f0 ) .
The work-energy theorem is then
R(1  cos f )(mg  f0 )  (1 / 2)mv 2f
(1)
d) When the object reaches the angle  f , apply Newton’s Second Law to determine
a second independent relationship between  f , m , g , R , and v f as needed.
Solution: when the object leaves the surface at  f , the normal force is zero, so Newton’s
Second Law in the radial direction is
mg cos f  mv 2f / R .
Thus
cos f  v 2f / gR
(2)
e) Using your results from parts c) and d), determine an expression for v f in terms
of m , g , R , and f 0 . Do not include  f in your answer.
Solution: Substitute Eq. (2) into Eq. (1) yields
R(1  v 2f / gR)(mg  f0 )  (1 / 2)mv 2f .
A little rearranging yields
R(mg  f0 )  (1 / 2)mv 2f  (v 2f / g)(mg  f0 )  (v 2f / g)((3 / 2)mg  f0 )
We can now solve for the speed of the object when it just leaves the sphere
v f  Rg
(mg  f0 )
.
((3 / 2)mg  f0 )
(3)