Question

Froghopper insects have a typical mass of around 12.7 mg and can jump to a height of 49.8 cm. The takeoff velocity is achieved as the insect flexes its legs over a distance of approximately 2.00 mm. Assume that the jump is vertical and that the froghopper undergoes constant acceleration while its feet are in contact with the ground. Ignore air resistance. What is the acceleration of the insect during the time of the jump (before it leaves the ground)?

Answer 1

Answer:

2403 m/s^2

Explanation:

The gravitational potential energy of the froghopper when it reaches the maximum height is equal to the kinetic energy at the moment of takeoff:

$mgh=\frac{1}{2}mv^2$

where

m = 12.7 mg is the mass of the froghopper

g = 9.8 m/s^2 is the acceleration due to gravity

h = 49.8 cm = 0.498 m is the maximum height reached

v = ? is the take-off velocity

Solving for v, we find

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(0.498 m)}=3.1 m/s$

We now that the froghopper accelerates from

u = 0 m/s

to

v = 3.1 m/s

in a distance of

d = 2.00 mm = 0.002 m

So we can find the acceleration by using the following SUVAT equation:

$v^2 - u^2 = 2ad$

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{(3.1 m/s)^2-0}{2(0.002 m)}=2403 m/s^2$