The froghopper (Philaenus spumarius), the champion leaper of the insect world, h

Question

The froghopper (Philaenus spumarius), the champion leaper of the insect world, has a mass of 12.3 mg. It can leave the ground with a speed as high as 4.0 m/s in the vertical direction. The jump itself lasts a mere 1.0 ms before the insect is clear of the ground. Assuming constant acceleration, find the force that the ground exerts on the froghopper during its jump.

What is the formula for the magnitude Fnet of the net force on the froghopper during the jump? Let m be the froghopper's mass and g be the magnitude of the acceleration due to gravity.

Now, use Newton's 2nd law F= ma to determine the magnitude of the normal force.

How many times larger than the insect's weight is this force?

Explanation / Answer

Mass of the frog hopper is m = 12.3 *10^-3 kg It can leave with a speed v = 4.0 m/s time is t = 1.0*10^-3 s a )  formula for the magnitude Fnet of the net force on the froghopper during the jump is       The magnitude of the normal force exerted by the ground is N = m ( a + g ) Fnet of the net force on the froghopper during the jump is F_net = N - mg b ) the insect's weight to the force is          N / w       m ( a + g ) / mg    = 12.3 *10^-3 kg ( 4.0 m/s / 1.0*10^-3 s ) + 9.8 m/s^2 / 12.3*10^-3 kg * 9.8 m/s^2 = 409 times

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