Q78P - University Physics with Modern Physics

Question

The Flying Leap of a Flea. High-speed motion pictures (3500 frames/second) of a jumping 210 - ug flea yielded the data to plot the flea’s acceleration as a function of time, as shown in Fig. P5.78. (See “The Flying Leap of the Flea,” by M. Rothschild et al., Scientific American, November 1973.) This flea was about  long and jumped at a nearly vertical takeoff angle. Using the graph, (a) find the initial net external force on the flea. How does it compare to the flea’s weight? (b) Find the maximum net external force on this jumping flea. When does this maximum force occur? (c) Use the graph to find the flea’s maximum speed.

                                 

Step-by-Step Solution

Answer

(a) The initial net external force on the flea is $1.28\times {10}^{-4}\mathrm{N}$ which is 62.5 times bigger than the flea’s weight.

(b) The maximum net external force on this jumping flea is $2.82\times {10}^{-4}\mathrm{N}$ and it occurs at $1.125\times {10}^{-3}\mathrm{s}$ .

(c) the maximum speed of flea is 1.51 m/s .

Step 1: Identification of the given data

 The given data can be listed below as:

• The mass of the flea is $\mathrm{m}=210\mathrm{\mu g}$ .
• The length of the flea is $\mathrm{l}=2 \mathrm{mm}\times \frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}=2\times {10}^{-3}\mathrm{m}$ .

Step 2: Significance of the force exerted

The force exerted is directly proportional to the mass and the acceleration of that object. Moreover, it is a vector quantity. The force is responsible for accelerating the body.

Step 3: (a) Determination of the initial net external force

According to the graph, the equation of the acceleration can be expressed as:

 

$$\begin{gathered} \frac{\mathrm{a}}{\mathrm{g}}=62.5 \\ \mathrm{a}=62.5\mathrm{g} \end{gathered}$$ 

 

Here, a is the acceleration of the flea and g is the acceleration due to gravity.

 

The equation of the initial net external force is expressed as:

 

 F = ma

Here, F is the initial net external force, m is the mass and is the acceleration of the flea.

 

Substitute the values in the above equation.

 

$$\begin{gathered} \mathrm{F}=\left(210\times {10}^{-9}\mathrm{kg}\right)\left(62.5\times 9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(210\times {10}^{-9}\mathrm{kg}\right)\left(612.5 \mathrm{m}/{\mathrm{s}}^2\right) \\ =1.28\times {10}^{-4}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =1.28\times {10}^{-4}\mathrm{N} \end{gathered}$$ 

 

Thus, the initial net external force on the flea is $1.28\times {10}^{-4}\mathrm{N}$ .

Step 4: (a) Comparison of the initial force with the flea’s weight

The equation of the flea’s weight is expressed as:

 

W = mg 

 

Here, W is the flea’s weight and g is the acceleration due to gravity. 

 

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{W}=\left(210\times {10}^{-9}\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =2.06\times {10}^{-6}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =2.06\times {10}^{-6}\mathrm{N} \end{gathered}$$

Thus, the initial net external force of the flea is  62.5 times bigger than the flea’s weight.

Step 5: (b) Determination of the maximum net external force

According to the graph, the equation of the maximum acceleration can be expressed as:

 

$$\begin{gathered} \frac{a_{max}}{g}=137.5 \\ {a}_{max}=137.5g \end{gathered}$$ 

 

Here, a is the acceleration of the flea and g is the acceleration due to gravity.

 

The equation of the initial net external force is expressed as:

 

 ${\mathrm{F}}_{\max }={\mathrm{ma}}_{\max }$

 

Here, ${\mathrm{F}}_{\max }$ is the initial net external force, m is the mass and $a_{\max }$ is the acceleration of the flea.

Substitute the values in the above equation.

 

$$\begin{gathered} {\mathrm{F}}_{\max }=\left(210\times {10}^{-9}\mathrm{kg}\right)\left(137.5\times 9.8 \mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(210\times {10}^{-9}\mathrm{kg}\right)\left(1347.5\mathrm{m}/{\mathrm{s}}^2\right) \\ =2.82\times {10}^{-4}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1 \mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2} \\ =2.82\times {10}^{-4}\mathrm{N} \end{gathered}$$ 

 

Thus, the maximum net external force on this jumping flea is $2.82\times {10}^{-4}\mathrm{N}$ .

Step 6: (b) Determination of the time

From the graph, the maximum force occurs at-

$$\begin{gathered} \mathrm{t}=1.125\mathrm{ms}\times \frac{{10}^{-3} \mathrm{s}}{1\mathrm{ms}} \\ =1.125\times {10}^{-3}\mathrm{s} \end{gathered}$$ .

Thus, the maximum force occurs at $1.125\times {10}^{-3}\mathrm{s}$ .

Step 7: (c) Determination of the maximum speed

The equation of the maximum speed of the flea is expressed as:

 

${\mathrm{F}}_{\max }=\mathrm{m}\frac{\mathrm{v}}{\mathrm{t}}$ 

 

Here, v  is the maximum speed.

 

Substitute the values in the above equation.

 

$$\begin{gathered} 2.82\times {10}^{-4}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2=\left(210\times {10}^{-9}\mathrm{kg}\right)\frac{\mathrm{v}}{1.125\times {10}^{-3}\mathrm{s}} \\ 2.82\times {10}^{-4}\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2=\left(1.86\times {10}^{-4}\mathrm{kg}/\mathrm{s}\right)\mathrm{v} \\ \mathrm{v}=1.51\mathrm{m}/\mathrm{s} \end{gathered}$$

 

Thus, the flea’s maximum speed is 1.51 m/s .