MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
IC_W07D2-1 Table Problem: Experiment 2 Cart-Spring on Inclined Plane Solution
An object of mass m slides down a plane that is inclined at an angle ! from the horizontal. The
object starts out at rest. The center of mass of the cart is a distance d from an unstretched spring
with spring constant k that lies at the bottom of the plane. Assume that the inclined plane has
a coefficient of kinetic friction µ . Find an equation whose solution describes how far the spring
compress will when the mass first comes to rest.
Solution:
Let x denote the displacement of the spring from the equilibrium position. Choose the zero
point for the gravitational potential energy U grav (x = 0) = 0 not at the very bottom of the inclined
plane, but at the location of the end of the unstretched spring. Choose the zero point for the
spring potential energy when the spring is at its equilibrium position, U spring (x = 0) = 0 .
a) Initial Energy: Choose for the initial state the instant the object is released. The initial kinetic
energy is K 0 = 0 . The initial potential energy is non-zero, U 0 = mg d sin ! . The initial
mechanical energy is then
E0 = K 0 + U 0 = mg d sin !
(1)
Final Energy: Choose for the final state the instant when the object first comes to rest and the
spring is compressed a distance x at the bottom of the inclined plane. The final kinetic energy is
K f = 0 since the mass is not in motion. The final potential energy is non-zero,
1 2
k x ! x mg sin " . Notice that the gravitational potential energy is negative because the
2
object has dropped below the height of the zero point of gravitational potential energy.
Uf =
The final mechanical energy is then
Ef = K f +U f =
1 2
k x ! x mg sin" .
2
(2)
Change in Mechanical Energy: The change in mechanical energy
$1
'
!Emechanical = E f " E0 = & k x 2 " x mg sin# ) " d mg sin# .
%2
(
(3)
The friction is primarily between the wheels and the bearings, not between the cart and the plane,
but the friction force may be modeled by a coefficient of friction µ . The effect of kinetic friction
is that there is now a non-zero non-conservative work done on the object, which has moved a
distance, d + x , given by
(
)
(
)
(
)
Wnc = ! f k d + x = ! µ N d + x = ! µ mg cos" d + x .
(4)
Note the normal force is found by using Newton’s Second Law in the direction
perpendicular to the inclined plane,
N ! mg cos" = 0 .
(5)
The Work-Energy Law,
Wnc = !Emechanical = E f " E0 ,
(6)
#1
&
! µ mg cos" d + x = % k x 2 ! x mg sin " ( ! d mg sin " .
$2
'
(7)
becomes
(
)
Equation (7) simplifies to
#1
&
0 = % k x 2 ! x mg sin " ! µ cos" ( ! d mg sin " ! µ cos" .
$2
'
(
)
(
)
(8)
This is a quadratic equation in x ,
x2 !
2mg (sin " ! µ cos" )
2mgd(sin " ! µ cos" )
x!
=0.
k
k
(9)
Although the problem does not specifically ask for us to solve the quadratic equation, we can use
the quadratic formula as long as we choose the positive choice of square root for the solution to
insure a positive displacement of the spring from equilibrium. The maximum displacement of the
spring is then
x=
.
#
&
mg
2k d
(sin! " µk cos! ) % 1 + 1 +
(.
k
mg(sin! " µk cos! ) '
$
(10)