Starting from rest, a 2.1x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts an average upward force of 0.37 N on it. This force does 1.5x10-4 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.
where did you get 0.0004076
Well, it sounds like this flea is really putting some effort into its takeoff. Let's break it down.
(a) To find the flea's speed when it leaves the ground, we can use the work-energy principle. The work done on the flea is equal to its change in kinetic energy. So, we have:
Work = ΔKE
Given that the work done on the flea is 1.5x10-4 J, we can set up the equation:
1.5x10-4 J = Δ(1/2 mv^2)
Where m is the mass of the flea (2.1x10-4 kg) and v is the unknown speed we're looking for.
Now, solving for v:
v^2 = (2(1.5x10-4 J))/(2.1x10-4 kg)
v^2 = 1.4286...
v ≈ √1.4286...
v ≈ 1.196 m/s
So, the flea's speed when it leaves the ground is approximately 1.196 m/s.
(b) To find how far upward the flea moves while pushing off, we need to calculate the displacement using the average force and work done. The work done is given as 1.5x10-4 J, and the average force is 0.37 N.
We know that work is equal to the force multiplied by the displacement:
1.5x10-4 J = 0.37 N * d
Solving for d:
d = (1.5x10-4 J) / (0.37 N)
d ≈ 0.000405 m
Therefore, the flea moves upward approximately 0.000405 meters while pushing off. Talk about a tiny hop!
Keep up the good work, little flea!
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.
(a) To find the flea's speed when it leaves the ground, we can use the work-energy principle.
The work done on the flea is given as 1.5x10^-4 J.
The work done on an object is equal to its change in kinetic energy:
Work = Change in Kinetic Energy
1.5x10^-4 J = (1/2)mv^2 - (1/2)m(0)^2 (since the flea starts from rest)
Where m is the mass of the flea and v is its final velocity.
The mass of the flea is given as 2.1x10^-4 kg.
1.5x10^-4 J = (1/2)(2.1x10^-4 kg)v^2
Simplifying the equation, we get:
1.5x10^-4 J = (1.05x10^-4 kg)v^2
Dividing both sides by 1.05x10^-4 kg, we get:
v^2 = 1.43 m^2/s^2
Taking the square root of both sides of the equation, we get:
v = 1.19 m/s
Therefore, the flea's speed when it leaves the ground is 1.19 m/s.
(b) To find the distance traveled by the flea while pushing off, we need to use the work-energy principle again.
The work done on the flea is given as 1.5x10^-4 J.
The work done on an object is equal to its change in kinetic energy:
Work = Change in Kinetic Energy
1.5x10^-4 J = (1/2)mv^2 - (1/2)m(0)^2 (since the flea starts from rest)
As the flea is pushing off vertically, the work done by the ground equals the work done on the flea.
1.5x10^-4 J = 0.37 N * h, where h is the height the flea rises.
Now, we can solve for h:
h = (1.5x10^-4 J) / (0.37 N)
h ≈ 0.405 m
Therefore, the flea moves approximately 0.405 meters upward while pushing off.
To solve this problem, we can apply the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
First, we need to determine the work done on the flea by the upward force from the ground. We are given that the work done is 1.5x10^(-4) J.
(a) To find the flea's speed when it leaves the ground, we can use the following equation:
Work = Change in kinetic energy
1.5x10^(-4) J = (1/2)mv^2
Where:
m = mass of the flea
v = velocity of the flea when it leaves the ground
We are given the mass of the flea (2.1x10^(-4) kg). Rearranging the equation, we can solve for v:
v^2 = (2 * 1.5x10^(-4) J) / (2.1x10^(-4) kg)
v^2 = (3x10^(-4) J) / (2.1x10^(-4) kg)
v^2 = 1.43 m^2/s^2
v = √(1.43) m/s
Therefore, the flea's speed when it leaves the ground is approximately 1.20 m/s.
(b) To find the distance upward traveled by the flea while it is pushing off, we need to determine the work done against gravity. Since there is no air resistance and the flea's weight is ignored, the work done against gravity is equal to the work done by the upward force from the ground.
The work done against gravity is equal to the force applied (0.37 N) multiplied by the distance traveled. Let's denote the distance traveled by the flea upward as d.
Work against gravity = Force x Distance
0.37 N x d = 1.5x10^(-4) J
Solve for d:
d = (1.5x10^(-4) J) / (0.37 N)
d = 4.05x10^(-4) m
Therefore, the flea moves approximately 4.05x10^(-4) meters upward while it is pushing off from the ground.
B)
W = Fd
(1.5 x 10^-4) = (0.37 - Fg)d
(1.5)(10^-4) = (0.37 - 9.8m)d
(1.5)(10^-4) = [0.37 - (9.8)(2.1)(10^-4)]d
0.00015 = 0.368d
d = 0.0004076 meters
A)
KE + PE = KE + PE
(1/2)mv^2 + (0) = (0) + mgh
(1/2)mv^2 = mgh
(1/2)v^2 = gh
v^2 = (2)(9.8)(0.0004076)
v = 0.08938 meters per second