1. Springs, Hooke’s Law, Work, and Energy
Chapter 4 in Book
4/14/2017
Dr. Sasho MacKenzie - HK 376

2. Objects that behave like springs
Pole vault pole
Golf club shaft
Tennis ball
Tennis racquet frame and strings
Hockey stick
A rubber band
4/14/2017
Dr. Sasho MacKenzie - HK 376

3. The Purpose of a Spring
Springs can be power amplifiers. Springs can return energy faster than the energy was stored in the spring.
Power is the rate at which work is done. P = W/t
Work done on an object changes the energy of the object by an amount equal to the work done.
Therefore, power is the rate at which energy is changing.
Power can also be calculated by multiplying the instantaneous net force acting on an object by the object’s instantaneous velocity. P = Fv
4/14/2017
Dr. Sasho MacKenzie - HK 376

4. Pole Vault Example: Store Energy
The vaulter applies force to the pole which causes the pole to bend
The more the pole bends, the more force is required to add additional bend.
The bend results in the storage of strain energy in the pole.
4/14/2017
Dr. Sasho MacKenzie - HK 376

5. Pole Vault: Return Energy
4. The vaulter then maneuvers into a position in which she does not supply enough resistance to maintain the bend in the pole.
The pole then releases the stored strain energy in the form of kinetic energy to the vaulter.
The correct timing of the return of energy from the pole and the fact that the energy is returned at a faster rate, enables the vaulter to jump higher.
4/14/2017
Dr. Sasho MacKenzie - HK 376

6. Hooke’s Law
If a ‘spring’ is bent, stretched or compressed from its equilibrium position, then it will exert a restoring force proportional to the amount it is bent, stretched or compressed.
Fs = kx (Hooke’s Law)
Related to Hooke’s Law, the work required to bend a spring is also a function of k and x.
Ws = (½ k)x2 = amount of stored strain energy
K is the stiffness of the spring. Bigger means Stiffer
x is the amount the spring is bent from equilibrium
4/14/2017
Dr. Sasho MacKenzie - HK 376

7. Based on Hooke’s Law A force is required to bend a spring
The more force applied the bigger the bend
The more bend, the more strain energy stored
Relationship between F and x
slope of the line = k
Force (N)
x (m)
Area under curve is the amount of stored strain energy
Stored strain energy = ½ k x2
4/14/2017
Dr. Sasho MacKenzie - HK 376

8. Kinetic Energy
When the strain energy stored in a spring is released, it is converted into kinetic energy.
The kinetic energy of an object is determined from an object’s mass and velocity. KE = ½ mv2
A 2 kg object moving at 3 m/s has a kinetic energy of…KE = ½(2)(3)2 = 9 J
Mass doesn’t change. This means that when strain energy is released, either the end of the spring or an object attached to it undergoes an increase in velocity.
The power is amplified.
4/14/2017
Dr. Sasho MacKenzie - HK 376

9. The Work-Energy Relationship
The work done on an object is equal to the net average force applied to the object multiplied by the distance over which the force acted. W = Fd
This means that no matter how long or how hard you apply a force to an object, no work is done if that object doesn’t move.
Mechanical work is different from physiological work.
100 J of work are done if a 100 N net force moves an object 1 m.
4/14/2017
Dr. Sasho MacKenzie - HK 376

10. Total Energy = KE + PE
An object’s total energy is calculated by adding its current kinetic (KE) and potential (PE) energies.
Potential energy is dependent upon an object’s current displacement from some reference point and a potential average force that will act on the object if it moves from its current position to that reference point.
An example of PE is an object at a certain height (h) above the ground. PE = mgh
A 4 kg object 2 m above the ground has PE = (9.81)(4)(2) = 78.5 J of potential energy.
4/14/2017
Dr. Sasho MacKenzie - HK 376

11. Total Energy = KE + PE
If the object is displaced from it’s current point to the reference point by that potential force, then it’s kinetic energy will increase by the amount of potential energy the object initially possessed.
A 4 kg object, initially at rest, that falls 2 m will have a KE = PE = (9.81)(4)(2) = 78.5 J.
Since KE = ½ mv2,…v = sqrt(2KE/m)
Therefore, the object will have a velocity of… v = sqrt[ 2(78.5)/4] = m/s when it lands.
4/14/2017
Dr. Sasho MacKenzie - HK 376

12. Potential Energy in a Spring
A second example of PE is stored strain energy.
Earlier it was shown that the strain energy stored in a linear spring is… SE = (½ k)x2, or ½Fx
In this case, the F stands for the amount of force required to deform the spring by its current displacement x.
The ½ is required for this PE equation because over the range of displacement, the force will fall to zero. Thus ½F represents the average force applied over the displacement.
4/14/2017
Dr. Sasho MacKenzie - HK 376

13. Strain PE vs. Gravitational PE
Force (N)
“mg”
x (m) “h”
Gravitational
Force (N)
x (m)
Strain
½Fx
Fx or
mgh
The area under the curve on the left equals the energy stored in a linear spring, or the amount of work required to deform the spring.
The area under the curve on the right equals the potential energy due to the constant force of gravity (mg), or the work required to lift an object x m.
Note that one area is square and the other triangular.
4/14/2017
Dr. Sasho MacKenzie - HK 376

14. Pole Vault Example
Yelena Isinbayeva (1.74m/65kg) deflects her 4.5m pole to 70% of it’s full length. The pole has a bending stiffness of 1000 N/m. At this point in the vault, her vertical velocity is 3 m/s and she is 2.5 m above the ground.
How much strain energy is stored in the pole?
What is her potential energy due to gravity?
How much kinetic energy does she have in the vertical direction?
What will be her peak height in the vault?
Dr. Sasho MacKenzie

15. Arampatzis et al., 2004
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16. Arampatzis et al., 2004
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17. Example: Slingshot
Jimmy loads a 2 kg water balloon into his giant slingshot which has a stiffness of 400 N/m. Jimmy stretches the slingshot 1.5 m from equilibrium and slings the balloon straight up into the air. If no strain energy is lost to heat or sound, how high will the balloon fly relative to the equilibrium point?
21.4 m
4/14/2017
Dr. Sasho MacKenzie - HK 376

18. A Tennis Serve During a tennis serve there are 3 springs in action
Frame 2. Ball Strings
Each spring has a specific stiffness (k) and therefore a different ability to act as a power amplifier.
The frame, the ball, and the strings all store and release energy differently
4/14/2017
Dr. Sasho MacKenzie - HK 376

19. The Strings x Stiff strings Force Compliant strings Load
The area under each line, is the amount of strain energy stored in the strings. You can see that less stiff springs (in this case strings) can store more strain energy for a given force.
However, the ball also has the ability to store and release strain energy. Perhaps the ball departs the strings before the ball has returned to equilibrium.
4/14/2017
Dr. Sasho MacKenzie - HK 376

20. Ball leaves before it (the ball) returns to equilibrium
Force x
This graph represents the ball only
Load
This strain energy will not increase the ball’s velocity.
Ball leaves at this displacement
This must be considered when the strings, frame and ball act as springs simultaneously during contact in a serve.
4/14/2017
Dr. Sasho MacKenzie - HK 376

21. x Frame Strings Force Ball: a player cannot alter the ball stiffness
The frame of a tennis racquet is a poor spring. Therefore, it is better to have the strings and ball performing most of the power amplification. Thus the frame should be the stiffest element.
Why would a good player still want stiff strings?
4/14/2017
Dr. Sasho MacKenzie - HK 376

22. Diving Board
How would you determine the actual stiffness of a diving board?
How does the stiffness of a diving board relate to the strain energy that can be stored temporarily in the board?
Sketch a vertical force-time profile for the force that acts on a diver’s feet during his time of contact with the board. Sketch the corresponding vertical force-time profile for the force from his feet pressing on the board.
4/14/2017
Dr. Sasho MacKenzie - HK 376

23. (Cross, 2000: Sport Eng)
4/14/2017
Dr. Sasho MacKenzie - HK 376