Chapter 10, Problem 10P

(a)

To determine

The average power required.

Answer to Problem 10P

The average power required is $1.2\times {10}^{-4}\text{ W}$ .

Explanation of Solution

Write the equation for the average power required.

$P_{\text{av}}=\frac{\Delta K}{\Delta t}$ (I)

Here, $P_{\text{av}}$ is the average power required, $\Delta K$ is the change in kinetic energy of the flee and $\Delta t$ is the time taken for the flee to reach its peak speed

Write the expression for $\Delta K$ .

$\Delta K=K_f-K_i$

Here, $K_f$ is the final kinetic energy of the flee and $K_i$ is the initial kinetic energy

The initial kinetic energy of the flee is zero.

Substitute $0$ for $K_i$ in the above equation.

$\begin{array}{c}\Delta K=K_f-0\\ \Delta K=K_f\end{array}$ (II)

Write the expression for $K_f$ .

$K_f=\frac{1}{2}m{v}_f^2$

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

Put the above equation in equation (II).

$\Delta K=\frac{1}{2}m{v}_f^2$

Put the above equation in equation (I).

$P_{\text{av}}=\frac{m{v}_f^2}{2\Delta t}$ (III)

Conclusion:

Given that the mass of the stunt flee is $0.45\times {10}^{-6}\text{ kg}$ its peak speed is $0.74\text{ m/s}$ and the time taken to reach the peak speed is $1.0\times {10}^{-3}\text{ s}$ .

Substitute $0.45\times {10}^{-6}\text{ kg}$ for $m$ , $0.74\text{ m/s}$ for $v_f$ and $1.0\times {10}^{-3}\text{ s}$ for $\Delta t$ in equation (III) to find the value of $P_{\text{av}}$ .

$\begin{array}{c}{P}_{\text{av}}=\frac{\left(0.45\times {10}^{-6}\text{ kg}\right){\left(0.74\text{ m/s}\right)}^2}{2\left(1.0\times {10}^{-3}\text{ s}\right)}\\ =1.2\times {10}^{-4}\text{ W}\end{array}$

Therefore, the average power required is $1.2\times {10}^{-4}\text{ W}$ .

(b)

To determine

Whether the insect muscle can provide the required power.

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.

$P_m=\left(0.20\right)m{P}_o$ (IV)

Here, $P_m$ is the power produced by the muscle and $P_o$ is the maximum power output of the muscle

Conclusion:

Given that the maximum power output of the muscle is $\left(60\text{ W/kg}\right)$ .

Substitute $0.45\times {10}^{-6}\text{ kg}$ for $m$ and $\left(60\text{ W/kg}\right)$ for $P_o$ in equation (IV) to find $P_m$.

$\begin{array}{c}{P}_m=\left(0.20\right)\left(0.45\times {10}^{-6}\text{ kg}\right)\left(\left(60\text{ W/kg}\right)\right)\\ =5.4\times {10}^{-6}\\ <P\end{array}$

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.

Answer to Problem 10P

The energy stored in the compression of the pads of the two hind legs is $3.7\times {10}^{-7}\text{ J}$ .

Explanation of Solution

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

$U=\frac{1}{2}k{\left(\Delta L\right)}^2$ (V)

Here, $U$ is the energy stored in a single pad, $k$ is the spring constant and $\Delta L$ 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.

$\begin{array}{c}E=2\left[\frac{1}{2}k{\left(\Delta L\right)}^2\right]\\ =k{\left(\Delta L\right)}^2\end{array}$ (VI)

Write the equation for $k$ .

$k=Y\frac{A}{L}$ (VII)

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

Write the equation for $A$ .

$A=L^2$

Put the above equation in equation (VI).

$\begin{array}{c}k=Y\frac{L^2}{L}\\ =YL\end{array}$

Put the above equation in equation (VI).

$E=YL{\left(\Delta L\right)}^2$

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

Replace $\Delta L$ in the above equation by $L$ .

$\begin{array}{c}E=YL{\left(L\right)}^2\\ =Y{L}^3\end{array}$ (VIII)

Conclusion:

Given that the value of the Young’s modulus is $1.7\text{ MPa}$ and the length of the side of the pad is $6.0\times {10}^{-5}\text{ m}$ .

Substitute $1.7\text{ MPa}$ for $Y$ and $6.0\times {10}^{-5}\text{ m}$ for $L$ in equation (VIII) to find $E$ .

$\begin{array}{c}E=\left(1.7\text{ MPa}\right){\left(6.0\times {10}^{-5}\text{ m}\right)}^3\\ =\left(1.7\times {10}^6\text{ Pa}\right){\left(6.0\times {10}^{-5}\text{ m}\right)}^3\\ =3.7\times {10}^{-7}\text{ J}\end{array}$

Therefore, the energy stored in the compression of the pads of the two hind legs is $3.7\times {10}^{-7}\text{ J}$ .

(d)

To determine

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

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.

$P_{\text{av}}=\frac{E}{\Delta t}$ (IX)

Conclusion:

Substitute $3.7\times {10}^{-7}\text{ J}$ for $E$ and $1.0\times {10}^{-3}\text{ s}$ for $\Delta t$ in equation (IX) to find $P_{\text{av}}$ .

$\begin{array}{c}{P}_{\text{av}}=\frac{3.7\times {10}^{-7}\text{ J}}{1.0\times {10}^{-3}\text{ s}}\\ =3.7\times {10}^{-4}\text{ W}\\ \text{>1}\text{.2}\times {10}^{-4}\text{ W}\end{array}$

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