Starting from rest, a 1.7x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.45 N on it. This force does 1.9x10-4 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.

I got (b) = 4.24x10-4 m but I can't figure out (a)

a. work= initial KE

work=1/2 m v^2 solve for v

But How to you solve for the velocity?

To find the flea's speed when it leaves the ground, you can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy:

Work = Change in kinetic energy

The work done on the flea is given as 1.9x10^-4 J, but we need to find its final kinetic energy.

The initial kinetic energy of the flea is zero, as it starts from rest. Therefore, the work done on the flea equals its final kinetic energy:

Work = Final kinetic energy

Let's represent the flea's final speed as v and its mass as m.

The work done on the flea is equal to the force applied (0.45 N) multiplied by the displacement (unknown). So we have:

Work = Force × Displacement

1.9x10^-4 J = 0.45 N × Displacement

To find the displacement, we need to use kinematic equations. The flea starts from rest, so its initial velocity u is zero.

We can use the equation of motion:

v^2 = u^2 + 2as

where v is the final velocity (unknown), u is the initial velocity (zero), a is the acceleration (unknown), and s is the displacement (unknown).

In this case, the acceleration is given by Newton's second law:

Force = mass × acceleration

0.45 N = (1.7x10^-4 kg) × acceleration

Now we have two equations:

1.9x10^-4 J = 0.45 N × Displacement
0.45 N = (1.7x10^-4 kg) × acceleration

The displacement and acceleration are related by the equation:

s = (v^2 - u^2) / (2a)

Substituting the values:

1.9x10^-4 J = 0.45 N × [(v^2 - 0) / (2 × (1.7x10^-4 kg × acceleration))]

Simplifying:

1.9x10^-4 J = 0.45 N × [(v^2) / (3.4x10^-4 (kg × acceleration))]

The units kg × acceleration in the denominator cancel out with N in the numerator, resulting in:

1.9x10^-4 J = (0.45 N / kg) × v^2

Rearranging the equation:

v^2 = (1.9x10^-4 J) / (0.45 N / kg)

Simplifying:

v^2 = (1.9x10^-4 J) / (0.45 N / kg)
v^2 = 4.22x10^-4 (J/N/kg)

Taking the square root of both sides yields:

v = √(4.22x10^-4 (J/N/kg))

Calculating this value gives:

v ≈ 0.0205 m/s

Therefore, the flea's speed when it leaves the ground is approximately 0.0205 m/s.

(Note: Please double-check the given values and calculations to ensure accuracy.)

To find the flea's speed when it leaves the ground, we need to use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, we know the work done on the flea (W = 1.9x10^(-4) J) and the average force exerted by the ground (F = 0.45 N).

(a) To find the flea's speed, we can use the formula for work done: W = ΔKE = (1/2)mv^2, where m is the mass of the flea and v is its speed.

First, let's find the mass of the flea:
Given: m = 1.7x10^(-4) kg

Now, let's rearrange the work formula to solve for speed (v):
W = (1/2)mv^2
v^2 = 2W/m
v = √(2W/m)

Plugging in the values:
v = √(2*(1.9x10^(-4))/(1.7x10^(-4)))

After evaluating, we find that the flea's speed when it leaves the ground is approximately 1.13 m/s.

For part (b), to find how far upward the flea moves while it is pushing off, we can use the work-energy principle again. The work done on the flea is equal to the change in potential energy.

(b) In this case, since the flea is moving vertically, the work done can be expressed as W = mgh, where m is the mass of the flea, g is the acceleration due to gravity, and h is the height the flea reaches.

Given: m = 1.7x10^(-4) kg
g = 9.8 m/s^2 (acceleration due to gravity)

Now, let's rearrange the formula to solve for height (h):
W = mgh
h = W/(mg)

Plugging in the values:
h = (1.9x10^(-4))/(1.7x10^(-4))*9.8

After evaluating, we find that the flea moves approximately 0.113 meters (or 4.24x10^(-4) meters) upward while pushing off the ground.