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5/1/2015Dr. Sasho MacKenzie - HK 3761 Springs, Hooke’s Law, Work, and Energy Chapter 4 in Book

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5/1/2015Dr. Sasho MacKenzie - HK 3762 Objects that behave like springs Pole vault polePole vault pole Golf club shaftGolf club shaft Tennis ballTennis ball Tennis racquet frame and stringsTennis racquet frame and strings Hockey stickHockey stick A rubber bandA rubber band

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5/1/2015Dr. Sasho MacKenzie - HK 3763 The Purpose of a Spring Springs can be power amplifiers. Springs can return energy faster than the energy was stored in the 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/tPower 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.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.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 = FvPower can also be calculated by multiplying the instantaneous net force acting on an object by the object’s instantaneous velocity. P = Fv

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5/1/2015Dr. Sasho MacKenzie - HK 3764 Pole Vault Example: Store Energy 1.The vaulter applies force to the pole which causes the pole to bend 2.The more the pole bends, the more force is required to add additional bend. 3.The bend results in the storage of strain energy in the pole.

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5/1/2015Dr. Sasho MacKenzie - HK 3765 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. 5.The pole then releases the stored strain energy in the form of kinetic energy to the vaulter. 6.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.

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5/1/2015Dr. Sasho MacKenzie - HK 3766 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.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. F s = k x (Hooke’s Law)F s = 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.Related to Hooke’s Law, the work required to bend a spring is also a function of k and x. W s = (½ k) x 2 = amount of stored strain energyW s = (½ k) x 2 = 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

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5/1/2015Dr. Sasho MacKenzie - HK 3767 Based on Hooke’s Law A force is required to bend a springA force is required to bend a spring The more force applied the bigger the bendThe more force applied the bigger the bend The more bend, the more strain energy storedThe more bend, the more strain energy stored Force (N) x (m) Relationship between F and x slope of the line = k Area under curve is the amount of stored strain energy Stored strain energy = ½ k x 2

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5/1/2015Dr. Sasho MacKenzie - HK 3768 Kinetic Energy When the strain energy stored in a spring is released, it is converted into 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 mv 2The kinetic energy of an object is determined from an object’s mass and velocity. KE = ½ mv 2 A 2 kg object moving at 3 m/s has a kinetic energy of…KE = ½(2)(3) 2 = 9 JA 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.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. The power is amplified.

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5/1/2015Dr. Sasho MacKenzie - HK 3769 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 dThe 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.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.Mechanical work is different from physiological work. 100 J of work are done if a 100 N net force moves an object 1 m.100 J of work are done if a 100 N net force moves an object 1 m.

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5/1/2015Dr. Sasho MacKenzie - HK 37610 Total Energy = KE + PE An object’s total energy is calculated by adding its current kinetic (KE) and potential (PE) energies.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.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 = mghAn 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.A 4 kg object 2 m above the ground has PE = (9.81)(4)(2) = 78.5 J of potential energy.

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5/1/2015Dr. Sasho MacKenzie - HK 37611 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.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.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 mv 2,…v = sqrt(2KE/m)Since KE = ½ mv 2,…v = sqrt(2KE/m) Therefore, the object will have a velocity of… v = sqrt[ 2(78.5)/4] = -6.3 m/s when it lands. Therefore, the object will have a velocity of… v = sqrt[ 2(78.5)/4] = -6.3 m/s when it lands.

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5/1/2015Dr. Sasho MacKenzie - HK 37612 Potential Energy in a Spring A second example of PE is stored strain energy.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) x 2, or ½F xEarlier it was shown that the strain energy stored in a linear spring is… SE = (½ k) x 2, or ½F x In this case, the F stands for the amount of force required to deform the spring by its current displacement 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.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.

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5/1/2015Dr. Sasho MacKenzie - HK 37613 Strain PE vs. Gravitational PE 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 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.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.Note that one area is square and the other triangular. Force (N) x (m) 0 Strain Force (N) “mg” x (m) “h” 0 Gravitational ½F x F x or mgh

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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.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. 1.How much strain energy is stored in the pole? 2.What is her potential energy due to gravity? 3.How much kinetic energy does she have in the vertical direction? 4.What will be her peak height in the vault? 14Dr. Sasho MacKenzie

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Arampatzis et al., 2004 15Dr. Sasho MacKenzie

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Arampatzis et al., 2004 16Dr. Sasho MacKenzie

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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?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? 5/1/2015Dr. Sasho MacKenzie - HK 37617

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5/1/2015Dr. Sasho MacKenzie - HK 37618 A Tennis Serve During a tennis serve there are 3 springs in actionDuring a tennis serve there are 3 springs in action 1.Frame2. Ball 3. Strings Each spring has a specific stiffness (k) and therefore a different ability to act as a power amplifier.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 differentlyThe frame, the ball, and the strings all store and release energy differently

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5/1/2015Dr. Sasho MacKenzie - HK 37619 The Strings Force xx Compliant strings Stiff strings 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. Load 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.

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5/1/2015Dr. Sasho MacKenzie - HK 37620 Ball leaves before it (the ball) returns to equilibrium Force xx Load Ball leaves at this displacement This strain energy will not increase the ball’s velocity. This must be considered when the strings, frame and ball act as springs simultaneously during contact in a serve. This graph represents the ball only

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5/1/2015Dr. Sasho MacKenzie - HK 37621 Force xx Ball: a player cannot alter the ball stiffness Frame Strings 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?

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5/1/2015Dr. Sasho MacKenzie - HK 37622 Diving Board 1.How would you determine the actual stiffness of a diving board? 2.How does the stiffness of a diving board relate to the strain energy that can be stored temporarily in the board? 3.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.

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5/1/2015Dr. Sasho MacKenzie - HK 37623 (Cross, 2000: Sport Eng)

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