Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10P

(a)

To determine

The average power required.

(a)

Expert Solution
Check Mark

Answer to Problem 10P

The average power required is 1.2×104 W .

Explanation of Solution

Write the equation for the average power required.

Pav=ΔKΔt (I)

Here, Pav is the average power required, ΔK is the change in kinetic energy of the flee and Δt is the time taken for the flee to reach its peak speed

Write the expression for ΔK .

ΔK=KfKi

Here, Kf is the final kinetic energy of the flee and Ki is the initial kinetic energy

The initial kinetic energy of the flee is zero.

Substitute 0 for Ki in the above equation.

ΔK=Kf0ΔK=Kf (II)

Write the expression for Kf .

Kf=12mvf2

Here, m is the mass of the flee and vf is its final speed

Put the above equation in equation (II).

ΔK=12mvf2

Put the above equation in equation (I).

Pav=mvf22Δt (III)

Conclusion:

Given that the mass of the stunt flee is 0.45×106 kg its peak speed is 0.74 m/s and the time taken to reach the peak speed is 1.0×103 s .

Substitute 0.45×106 kg for m , 0.74 m/s for vf and 1.0×103 s for Δt in equation (III) to find the value of Pav .

Pav=(0.45×106 kg)(0.74 m/s)22(1.0×103 s)=1.2×104 W

Therefore, the average power required is 1.2×104 W .

(b)

To determine

Whether the insect muscle can provide the required power.

(b)

Expert Solution
Check Mark

Answer to Problem 10P

The flea’s muscle cannot provide the power required.

Explanation of Solution

Given that only 20% of the flea’s weight is muscle.

Write the equation for the power produced by the muscle.

Pm=(0.20)mPo (IV)

Here, Pm is the power produced by the muscle and Po is the maximum power output of the muscle

Conclusion:

Given that the maximum power output of the muscle is (60 W/kg) .

Substitute 0.45×106 kg for m and (60 W/kg) for Po in equation (IV) to find Pm.

Pm=(0.20)(0.45×106 kg)((60 W/kg))=5.4×106<P

The power produced by the muscle is smaller than the power required.

Therefore, the flea’s muscle cannot provide the power required.

(c)

To determine

The energy stored in the compression of the pads of the two hind legs.

(c)

Expert Solution
Check Mark

Answer to Problem 10P

The energy stored in the compression of the pads of the two hind legs is 3.7×107 J .

Explanation of Solution

Write the equation for the energy stored in a single pad.

U=12k(ΔL)2 (V)

Here, U is the energy stored in a single pad, k is the spring constant and ΔL is the length change

There are two pads.

Write the equation for the total energy stored.

E=2U

Here, E is the total energy stored

Put equation (V) in the above equation.

E=2[12k(ΔL)2]=k(ΔL)2 (VI)

Write the equation for k .

k=YAL (VII)

Here, Y is the Young’s modulus, A is the cross sectional area and L is the initial length

Write the equation for A .

A=L2

Put the above equation in equation (VI).

k=YL2L=YL

Put the above equation in equation (VI).

E=YL(ΔL)2

Given that the pad compresses fully so that the change in length is equal to the initial length.

Replace ΔL in the above equation by L .

E=YL(L)2=YL3 (VIII)

Conclusion:

Given that the value of the Young’s modulus is 1.7 MPa and the length of the side of the pad is 6.0×105 m .

Substitute 1.7 MPa for Y and 6.0×105 m for L in equation (VIII) to find E .

E=(1.7 MPa)(6.0×105 m)3=(1.7×106 Pa)(6.0×105 m)3=3.7×107 J

Therefore, the energy stored in the compression of the pads of the two hind legs is 3.7×107 J .

(d)

To determine

Whether the resilin pads .provide enough power for the jump.

(d)

Expert Solution
Check Mark

Answer to Problem 10P

The resilin pads .provide enough power for the jump.

Explanation of Solution

Write the equation for the average power provided by the resilin pads.

Pav=EΔt (IX)

Conclusion:

Substitute 3.7×107 J for E and 1.0×103 s for Δt in equation (IX) to find Pav .

Pav=3.7×107 J1.0×103 s=3.7×104 W>1.2×104 W

Therefore, the resilin pads .provide enough power for the jump.

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Chapter 10 Solutions

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