- #1

pepov

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## Homework Statement

High-speed motion pictures

**(3500 frames/second)**of a jumping 240μg flea yielded the data to plot the flea's acceleration as a function of time as shown in the figure (Figure 1) . (See "The Flying Leap of the Flea," by M. Rothschild et al. in the November 1973 Scientific American.) This flea was about 2 mm long and jumped at a nearly vertical takeoff angle. Use the measurements shown on the graph to answer the questions.

1) Find the

*initial*net external force on the flea

2) How does it compare to the flea's weight

3) Find the

*maximum*net external force on this jumping flea

4) When does this maximum force occur?

5) Use the graph to find the flea's maximum speed.

## Homework Equations

F = ma

a = Δv/Δt

## The Attempt at a Solution

1)

m = .24 mg

I began by looking up the article because I did not understand what a/g on the y-axis means. I found that this is "Accerleration (Gravities)". It looks to me like the graph begins at ~60 a/g so the force of the jump is:

a

_{j}= 60*9.81 = 588.6 m/s

^{2}

F

_{j}= m*a

_{j}= 588.6 m/s

^{2}*.24mg = 141.26 N

This is the force in the vertical direction. So I added the force of the flea's weight:

F

_{w}= .24mg * -9.81 m/s

^{2}

F

_{w}= -2.35 N

F

_{w}+ F

_{j}= F

_{net}

-2.35 N + 141.26 N = 138.91 N

This answer is incorrect. Perhaps the initial force should only include the weight of the flea? Although this wouldn't make sense to me because then the graph would be irrelevant.

2)

Given the answer to (1) I should divide by F

_{w}:

[Answer to (1)] / 2.35 = __ w

3)

The

*maximum*net external force would be mass times the flea's highest a/g which looks to be around 140 a/g.

140 a/g * 9.81 = 1373.4 m/s

^{2}

F

_{max}= 1373.4 m/s

^{2}*.24mg

F

_{max}= 329.616 N

Following the same procedure as before (which will change when I figure out what I did wrong):

F

_{max}+ F

_{w}= F

_{netMAX}

329.616 - 2.35 = 327.266 N

This is also incorrect.

4)

I will figure this out when I figure out 3 by looking at the graph.

5)

This was easy and I just looked at the graph for the answer.

Thanks very much for all your help,

Pepov

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