EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684651
Author: Katz
Publisher: VST
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Chapter 8, Problem 32PQ

The Flybar high-tech pogo stick is advertised as being capable of launching jumpers up to 6 ft. The ad says that the minimum weight of a jumper is 120 lb and the maximum weight is 250 lb. It also says that the pogo stick uses “a patented system of elastometric rubber springs that provides up to 1200 lbs of thrust, something common helical spring sticks simply cannot achieve (rubber has 10 times the energy storing capability of steel).”

Chapter 8, Problem 32PQ, The Flybar high-tech pogo stick is advertised as being capable of launching jumpers up to 6 ft. The

  1. a. Use Figure P8.32 to estimate the maximum compression of the pogo stick’s spring. Include the uncertainty in your estimate.
  2. b. What is the effective spring constant of the elastometric rubber springs? Comment on the claim that rubber has 10 times the energy-storing capability of steel.
  3. c. Check the ad’s claim that the maximum height a jumper can achieve is 6 ft.

(a)

Expert Solution
Check Mark
To determine

The maximum compression of the pogo sticks spring with uncertainty.

Answer to Problem 32PQ

The maximum compression of the pogo sticks spring is 0.3m±0.1m_.

Explanation of Solution

The maximum compression of the spring will be always less than the length of the stick below the person’s feet. It is estimated that the maximum compression l to be 1 foot or about 0.3m.

The uncertainty of the estimation will be about ±0.1m.

Conclusion:

Therefore, the maximum compression of the pogo sticks spring is 0.3m±0.1m_.

(b)

Expert Solution
Check Mark
To determine

The effective spring constant of the elastometric rubber springs and comment on the claim that rubber has 10 times the energy storing capability of steel.

Answer to Problem 32PQ

The effective spring constant of the elastometric rubber springs is 1800N/m_ and it does not make sense to say from this example that rubber spring stores more energy than a steel spring because it depends on the geometric parameters of the spring.

Explanation of Solution

Write the expression force exerted by the spring when it is fully compressed according to Hooke’s law.

    FH=kl                                                                                                                   (I)

Here, FH is the force exerted by the spring, k is the spring constant, l is the maximum compression.

The effective spring constant must be derived from the claim that the spring provides 1,200lbs of thrust which we assume is the force exerted by the spring when it is fully compressed.

Solve equation (I) for k,

    k=FHl                                                                                                                    (II)

Conclusion:

Substitute 1200lbs for FH, 0.3m for l in equation (II) to find k.

    k=FHl=1200lbs0.3m=1200lbs×4.45N1lbs0.3m1800N/m

Since this is a pretty stiff spring, it doesn’t make sense to say that the rubber spring stores more energy than steel spring because it depends on the geometric c parameters of the spring.

Therefore, the effective spring constant of the elastometric rubber springs is 1800N/m_ and it is not possible to say from this example that rubber spring stores more energy than a steel spring because it depends on the geometric parameters of the spring

(c)

Expert Solution
Check Mark
To determine

Show that the maximum height a jumper can achieve is 6ft.

Answer to Problem 32PQ

The result is far below the claimed maximum height of 6ft.

Explanation of Solution

The total energy at the two points in the trajectory (1) the jumper is momentarily at rest on the ground before the jump with zero gravitational potential energy and the spring is fully compressed (II) is the jumper is momentarily at rest at peak height when all of the energy is in gravitational potential energy.

Write the expression for the conservation of the total mechanical energy.

    ΔK+ΔU=0                                                                                                       (III)

Here, ΔK is the change in kinetic energy, ΔU is the change in potential energy.

The change in kinetic energy between the two points is zero, the change in potential energy is,

    ΔU=ΔUg+ΔUe                                                                                               (IV)

Here, ΔUg is the change in gravitational potential energy, ΔUe is the change in elastic potential energy.

Write the expression for the change in gravitational potential energy.

    ΔUg=mgh                                                                                                           (V)

Here, m is the mass of the body, g is the acceleration due to gravity, h is the height of the body from ground.

Write the expression for the change in elastic potential energy.

    ΔUe=12kl2                                                                                                          (VI)

Use equation (IV), (V) and (VI) in equation (III) and solve for h.

    0+mgh12kl2=0h=kl22mg                                                                                    (VII)

Conclusion:

Substitute 1800N/m for k, 0.3m for l, 60kg for m, 9.81m/s2 for g in equation (VII) to find h.

    h=(1800N/m)(0.3m)22(60kg)(9.81m/s2)=0.14m×3.280ft1m=0.5ft

The obtained result is far below the claimed maximum height of 6 ft. The biggest uncertainty comes from the statement about “thrust then the spring constant must be about ten times larger.

Therefore, the result is far below the claimed maximum height of 6ft.

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Chapter 8 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

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