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What is the flea’s Kinetic Energy as it leaves the ground? A $0.50 mg$ flea, jumping straight up, reach a height of $30 cm$ if there were no air resistance. In reality, air resistance limits the height to $20 cm$.

The question aims to calculate the kinetic energy of a flea whose mass is $0.50 mg$ and has attained the height of $30 cm$, provided that there is no air resistance.

The kinetic energy of an object is defined as the energy it has acquired due to its motion. In other terms, this can also be defined as the work done to move or accelerate an object of any mass from rest to any position with the desired or set velocity. The kinetic energy gained by the body remains the same until the velocity remains constant during the course of its movement.

The formula for Kinetic energy is given as:

\[ K.E = 0.5mv^2 \]

Air resistance is referred to as opposing forces that oppose or restrict the motion of the objects as they move through the air. Air resistance is also called as drag force. Drag is a force that acts on an object in the opposite direction of its travel. It’s been said to be “the greatest killer” because it has this amazing power not just for stopping but also for accelerating motion.

In this case, air resistance has been ignored.

Expert Answer:

In order to find out the Kinetic Energy of the flea, lets first calculate its initial velocity using the following second equation of motion:

\[ 2aS = (v_f)^2 – (v_i)^2 \]

Where:

$a$ is gravitational acceleration that is equivalent to $9.8 m/s^2$.

$S$ is the height without considering the effect of air resistance, given as $30 cm = 0.30 m$

$v_f$ is the final velocity of the flea which is equivalent to $0$.

Let’s put the values in the equation to calculate the initial velocity $v_i$.

\[ 2(9.8)(0.30) = (0)^2 – (v_i)^2 \]

\[ (v_i)^2 = 5.88   \]

\[ v_i = 2.42   m/s^2 \]

Now let’s calculate the Kinetic energy by using the following equation:

\[ K.E = 0.5mv^2 \]

Where $m$ is the mass, given as $0.5 mg = 0.5\times{10^{-6}} kg$.

\[ K.E = 0.5(0.5\times{10^{-6}})(2.42)^2 \]

\[ K.E = 1.46\times{10^{-6}} J \]

Therefore, the Kinetic Energy of the flea as it leaves the ground is given as $1.46\times{10^{-6}} J$.

Alternative Solution:

This question can also be solved by using the following method.

Kinetic Energy is given as:

\[ K.E = 0.5mv^2 \]

Whereas the Potential Energy is given as:

\[ P.E = mgh \]

Where $m$ = mass, $g$ = gravitation acceleration and $h$ is height.

Let’s first calculate the flea’s Potential Energy.

Substituting values:

\[ P.E = (0.5\times{10^{-6}})(9.8)(0.30) \]

\[ P.E = 1.46\times{10^{-6}} J \]

According to the law of conservation of energy, the potential energy at the top is exactly similar to kinetic energy at the ground.

So:

\[ K.E = P.E \]

\[ K.E = 1.46\times{10^{-6}} J \]

Example:

Fleas have a remarkable jumping ability. A $0.60 mg$ flea, jumping straight up, would reach a height of $40 cm$ if there were no air resistance. In reality, air resistance limits the height to $20 cm$.

  1. What is the flea’s potential energy at the top?
  2. What is the flea’s kinetic energy as it leaves the ground?

Given these values:

\[ m = 0.60 mg = 0.6\times{10^{-6}}kg \]

\[ h = 40 cm = 40\times{10^{-2}}m = 0.4 m \]

1) Potential Energy is given as:

\[ P.E = mgh \]

\[ P.E = (0.6\times{10^{-6}})(9.8)(0.4) \]

\[ P.E = 2.35\times{10^{-6}} \]

2) According to the law of conservation of energy,

Kinetic energy at the ground = Potential energy at the top

So:

\[ K.E = 2.35\times{10^{-6}} \]

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