Horizontal Mass Spring System

WORK DONE BY A SPRING

MINGXUAN LIU

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.

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

A interesting problem showing how energy is

convertedSuppose 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?

Solution to the problem

16m

3m

2.7m0.3m

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

Solution to the problem

16m

3m

2.7m0.3m

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=0.5kx22

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)(Remember that hmax is the distance above the compressed spring, the absolute height is 15.18m which is > 13m)

FLOOR

CEILING

A

B

Increase Height

Change the sign!

hmax

Solution to the problem

16m

3m

2.7m0.3m

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 x be the distance spring B is compressed.

FLOOR

CEILING

A

B

original

x

h

Solution to the problem

16m

3m

2.7m0.3m

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.It’s not fair for the ball to lose so much energy, spring B which is compressed by the ball must take that energy to maintain the balance.

FLOOR

CEILING

A

B

original

x

h=12.7

Solution to the problem

16m

3m

2.7m0.3m

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

original

x

h=12.7

The end

Thanks for watching.