A force platform is a tool used to analyze the performance of athletes measuring the vertical force that the athlete exerts on the ground as a function of time. Starting from rest, a 60.0 kg athlete jumps down onto the platform from a height of 0.600 m. While she is in contact with the platform during the time interval 0 < t < 0.8 s, the force she exerts on it is described by the function below.
F = (9 200 N/s)t - (11 500 N/s2)t2
Assume the positive y-axis points upward.
(a) What was the athlete's velocity when she reached the platform?
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(b) What impulse did the athlete receive from the platform?
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(c) What impulse did the athlete receive from gravity while in contact with the platform?
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(d) With what velocity did she leave the platform?
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Your response differs from the correct answer by more than 100%. j hat
(e) To what height did she jump upon leaving the platform?
To find the answers to parts (c), (d), and (e) of the question, we need to first calculate the impulse received from gravity while in contact with the platform.
Impulse is defined as the change in momentum and is equal to the integral of force with respect to time.
(c) To find the impulse received from gravity, we need to integrate the force due to gravity over the time interval during which the athlete is in contact with the platform.
The force due to gravity is given by F = mg, where m is the mass of the athlete and g is the acceleration due to gravity.
In this case, m = 60.0 kg and g = 9.8 m/s^2.
So the force due to gravity is F = (60.0 kg) * (9.8 m/s^2) = 588 N.
The impulse received from gravity can now be calculated by integrating this force over the time interval 0 < t < 0.8 s.
∫[0,0.8] (588 N) dt = (588 N) * t |[0,0.8] = (588 N) * (0.8) - (588 N) * (0) = 470.4 Ns.
Therefore, the athlete received an impulse of 470.4 Ns from gravity while in contact with the platform.
(d) To find the velocity with which she left the platform, we can apply the impulse-momentum principle, which states that the impulse received by an object is equal to the change in its momentum.
The impulse received from the platform will be equal in magnitude and opposite in direction to the impulse received from gravity. So the impulse received from the platform is -470.4 Ns.
The velocity with which she left the platform can be calculated using the formula:
Impulse = (mass of the athlete) * (change in velocity)
So we have -470.4 Ns = (60.0 kg) * (change in velocity)
Solving for the change in velocity:
change in velocity = -470.4 Ns / (60.0 kg) = -7.84 m/s
Therefore, she left the platform with a velocity of -7.84 m/s (negative sign indicates direction).
(e) To find the height to which she jumped upon leaving the platform, we can use the conservation of mechanical energy.
The mechanical energy at the start, when she was at rest on the platform, is given by the potential energy:
Potential energy = m * g * h
where m is the mass of the athlete, g is the acceleration due to gravity, and h is the initial height.
Potential energy at the start = (60.0 kg) * (9.8 m/s^2) * (0.600 m) = 352.8 J
The mechanical energy at the end, when she reaches the maximum height, is given by the sum of her potential energy and kinetic energy:
Mechanical energy at the end = Potential energy at the end + Kinetic energy at the end
Since she reaches the maximum height and then falls back down, her velocity at the maximum height is 0 m/s.
So the mechanical energy at the end is equal to the potential energy at the maximum height:
Mechanical energy at the end = (60.0 kg) * (9.8 m/s^2) * (height at the end)
Now we can set up the conservation of mechanical energy equation:
Potential energy at the start = Mechanical energy at the end
352.8 J = (60.0 kg) * (9.8 m/s^2) * (height at the end)
Solving for the height at the end:
height at the end = 352.8 J / ((60.0 kg) * (9.8 m/s^2)) = 0.600 m
Therefore, the athlete jumped back up to the same height of 0.600 m upon leaving the platform.