Starting from rest, a 1.6x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.38 N on it. This force does 2.6x10-4 J of work on the flea. (a) What is the flea's speed when it leaves the ground? (b) How far upward does the flea move while it is pushing off? Ignore both air resistance and the flea's weight.

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To solve this problem, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. We can use this principle to find the velocity of the flea when it leaves the ground.

(a) First, let's calculate the work done on the flea by using the given force and the distance over which it acts:

Work (W) = Force (F) * Distance (d)

Given:
Force (F) = 0.38 N
Distance (d) = Unknown

The work done on the flea is equal to the change in its kinetic energy:

Work (W) = ΔKE

We know that the initial kinetic energy (KEi) is zero because the flea starts from rest. Therefore, we can rewrite the equation as:

ΔKE = KEf - KEi

Since ΔKE = Work (W), we have:

W = KEf - KEi

The final kinetic energy (KEf) is given by:

KEf = (1/2) * m * v^2

Where:
m = mass of the flea = 1.6x10^-4 kg (given)
v = velocity of the flea

Now, substituting the values into the equation:

0.38 N * d = (1/2) * 1.6x10^-4 kg * v^2

Simplifying the equation, we get:

0.38 N * d = 8x10^-5 kg * v^2

Dividing both sides by 8x10^-5 kg:

(0.38 N * d) / (8x10^-5 kg) = v^2

Taking the square root of both sides to solve for v:

v = √[(0.38 N * d) / (8x10^-5 kg)]

(b) To find the distance upward the flea moves while pushing off, we can use the work-energy principle again. The work done by the ground on the flea is equal to the flea's initial kinetic energy:

Work (W) = KEi

Given:
W = 2.6x10^-4 J
KEi = 0 (since the flea starts from rest)

Therefore:

2.6x10^-4 J = 0.5 * 1.6x10^-4 kg * v^2

Simplifying the equation:

5.2 * 10^-4 J = 1.6x10^-4 kg * v^2

Dividing both sides by 1.6x10^-4 kg:

v^2 = (5.2 * 10^-4 J) / (1.6x10^-4 kg)

Taking the square root of both sides to find v:

v = √[(5.2 * 10^-4 J) / (1.6x10^-4 kg)]

Now, we can calculate both the speed of the flea when it leaves the ground (a) and the distance it moves upward while pushing off (b) using the formulas we derived above.

To answer this question, we'll need to use the concepts of work and energy. We can solve both parts of the question by applying the work-energy principle.

a) First, let's determine the flea's speed when it leaves the ground.
The work done on an object is equal to the change in its kinetic energy, so we can write:

Work = ΔKE

Given: Work = 2.6x10^(-4) J

The flea starts from rest, so initially, its kinetic energy is zero (KE_initial = 0).

Therefore, we can rewrite the equation as:

2.6x10^(-4) J = KE_final - KE_initial

Since KE_initial is zero, we have:

2.6x10^(-4) J = KE_final

Next, we can use the equation for kinetic energy:

KE = (1/2)mv^2

where KE is the final kinetic energy, m is the mass of the flea, and v is the speed.

Rearranging the equation, we find:

v^2 = (2KE) / m

Substituting the given values, we get:

v^2 = (2 * 2.6x10^(-4) J) / (1.6x10^(-4) kg)

v^2 = 2.6 J / 0.16 kg

v^2 = 16.25 m^2/s^2

Taking the square root of both sides, we find:

v ≈ 4.03 m/s

Therefore, the flea's speed when it leaves the ground is approximately 4.03 m/s.

b) Now let's determine how far upward the flea moves while it is pushing off.

The work done on the flea is given by the equation:

Work = force * distance

Given: Work = 2.6x10^(-4) J
Force = 0.38 N

We need to find the distance.

Rearranging the equation for work, we get:

distance = Work / force

Substituting the values, we have:

distance = (2.6x10^(-4) J) / (0.38 N)

distance = 6.842x10^(-4)‬ m

Therefore, the flea moves approximately 6.842x10^(-4)‬ m upward while it is pushing off.

Note: It's important to mention that this solution assumes ideal conditions, neglecting air resistance and the flea's weight, as stated in the question.