work done by a spring

1. Horizontal Mass Spring
System
WORK DONE BY A SPRING
MINGXUAN LIU

2. Some facts about a
spring
A spring always keeps its original length.
The force exerted by a spring is proportional to
its stretched/compressed length.
Thus, the force for holding a spring is not a
constant but can be calculated using Hooke’s law.
Where F is the force exerted by the spring, k is
the spring constant showing how much the force
increase with stretched/compressed length, x is
the stretched/compressed length.

3. Work done by a spring
In high school we known that the work done by
something equal to the product of the exerted
force and moving distance.
Applying the work equation and the force exerted
by a spring, the work done by a spring is:
After evaluating this integral we have:
Where x1 is the initial position,
and x2 is the final position.
Images from Physics For Scientists And
Engineers – 1e

4. A interesting problem showing
how energy is converted
Suppose there is a tower that is 16 meters tall. Two
identical spring A, B of natural length 3m and spring
constant 200N/m are assembled on the floor and
ceiling. Both of them point toward each other directly.
A metal ball of mass 5kg is placed on the bottom
spring(A) and pushed by its maximum force (by which
the spring is compressed to 10% of its natural length)
of the spring. Ignoring the air resistance and energy
losing, assume that the ball moves perpendicular to
the ground without any angle: After the spring is
released, will the ball hitting the upper spring (B) on
the ceiling? If so, how much (as much as possible)
will the spring (B) be compressed in meters?

5. Solution to the problem
16m
3m
2.7m
0.3m
5kg
200N/m
200N/m
Step1: Sketch the picture with all the key
information labeled on it. From the picture we know
that since the floor and ceiling are fixed, when
spring A is released, the ball moves upward and
may hit the upper spring.
Step2: Figure out how the energy converts among
the ball, spring A and spring B.
When spring A is released:
Work done by A = Kinetic plus gravitational potential
energy of the ball = Gravitational potential energy
of the ball when reach the highest possible height
If you are confused think about that if the ceiling is
high enough and at the time the velocity of the up
moving ball decrease to 0, it reaches the highest
point. Then fall down.FLOOR
CEILING
A
B

6. Solution to the problem
16m
3m
2.7m
0.3m
5kg
200N/m
200N/m
Gravitational potential energy of the ball at the highest
possible height is given by: Ep=mballghmax
We know that all of the gravitational potential energy
comes from the spring.
Espring Ek+Ep Ep
The work done by spring A is given by: Ws=-0.5kx2
2
(x1=0)
Combine two equation we have: -Ep=Ws
mballghmax=0.5kx2
2
5.0kg*9.8m/s2*hmax=0.5*200N/m*(0.9*3m)2
hmax=
(0.5*200N/m*(0.9*3m)2)/(5.0kg*9.8m/s2)
hmax = 14.88m (hit the upper spring)
FLOOR
CEILING
A
B
Increase Height
Change the
sign!
hmax

7. Solution to the problem
16m
3m
2.7m
0.3m
5kg
200N/m
200N/m
Step3: Now we know that the ball will hit spring B,
but how much will the spring be compressed?
Consider that the total energy in this system doesn’t
change. If something loses its energy, something
must gain that amount of energy. When the ball is
on its way up, work done by spring A convert into
both EK and Ep of the ball. Also, Ek goes into Ep at
the same time by lowering the upward velocity.
After the ball contact with spring B, part of its Ek
goes into the potential energy of spring B and part
of its Ek goes into Ep. AND Ek IS TOTALLY
CONVERTED WHEN SPRING B IS
COMPRESSED AS MUCH AS POSSIBLE. When
this point is reached, the ball is at a rest, all of it
energy takes form of Ep. Let h be the distance
between where spring A is compressed and the
point where the ball is almost touching spring B. Let
FLOOR
CEILING
A
B
origina
l
x
h

8. Solution to the problem
16m
3m
2.7m
0.3m
5kg
200N/m
200N/m
Step4: Imaging if there is no upper spring, the ball
can go as high as about 15m. However, the
presence of spring B trapped the ball and absorb
some of its energy.
Without the ceiling, the ball at max height has
energy: EP=mballghmax (assume that +0.3m is the
ground level)
With the ceiling, the ball at max height has energy:
EP=mballg(h+x)
THUS, THE ENERGY DIFFERENCE IN THESE
TWO SITUATIONS SHOULD EQUAL TO THE
ENERGY ABSORBED BY SPRING B.
If you are still confused, consider that (h+x) is a
smaller number than hmax, the ball with hmax has a
larger energy.
FLOOR
CEILING
A
B
origina
l
x
h=12.7

9. Solution to the problem
16m
3m
2.7m
0.3m
5kg
200N/m
200N/m
Combine the equation we get:
EspringB = -WspringB= mballghmax -mballg(h+x)
0.5*200N/m*x2=5.0kg*9.8m/s2(14.88m-x-
12.70m)
100x2=49(2.18-x)
100x2+49x-106.82=0
x=0.82m (discard the negative
value)
Conclusion: The ball will hit the upper spring and
cause a compress of 0.82m to it. The key part of
this problem is realizing how the energy is
converted in the spring system. Once you master
where the energy goes and build a energy balance
equation, the reaming part is just doing algebra.
FLOOR
CEILING
A
B
origina
l
x
h=12.
7

10. The end
Thanks for watching.