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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)?

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skyluke89

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:

[tex]mgh=\frac{1}{2}mv^2[/tex]

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

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

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:

[tex]v^2 - u^2 = 2ad[/tex]

Solving for a,

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

The acceleration of the Froghopper insect during the time of the jump (before it leaves the ground) is 2403 m/s².

What is energy?

Energy is the capacity of a object to do a work. More the energy object posses, more the work it can do.

The main categories of energy

  • 1. Kinetic—The energy of moving objects. Kinetic energy is half of the product of mass and squared velocity.
  • 2. Potential—The energy stored in objects. Potential energy due to gravity is the product of mass, gravitational force and height of the body.

The kinetic energy of the Frog hopper at the time of takeoff is equal to the potential energy due to gravity at the maximum height.

[tex]\dfrac{1}{2}mv^2=mgh\\v^2=2gh\\v=\sqrt{2gh}[/tex]

Here, (v) is the velocity, and (h) is the maximum height.

The frog hopper insects can jump to a height of 49.8 cm or 0.498 m and the value of gravitational force is 9.8 m/s². Plug in the value in above expression as,

[tex]v=\sqrt{2(9.8)(0.498)}\\\v=3.1\rm\; m/s[/tex]

The takeoff velocity is achieved as the insect flexes its legs over a distance of approximately 2.00 mm or 0.002 m and the initial velocity of the zero. Thus, by the equation of motion,

[tex]v^2=u^2+2as\\(3.1)^2=0+2a(0.002)\\a=2403\rm \; m/s^2[/tex]

Thus, the acceleration of the Froghopper insect during the time of the jump (before it leaves the ground) is 2403 m/s².

Learn more about the energy here;

https://brainly.com/question/8101588

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