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?
(b) What impulse did the athlete receive from the platform?
(c) What impulse did the athlete receive from gravity while in contact with the platform?
(d) With what velocity did she leave the platform?
(e) To what height did she jump upon leaving the platform?
To solve the given problems, we need to use the equations of motion and the concept of impulse.
(a) To find the athlete's velocity when she reached the platform, we need to calculate the integral of the force function with respect to time, and then solve for velocity.
Step 1: Calculating the integral of the force function:
∫F dt = ∫[(9,200 N/s)t - (11,500 N/s^2)t^2] dt
Integrating the first term, we get:
= (9,200 N/s) * 0.5t^2 - (11,500 N/s^2) * (1/3)t^3
Step 2: Solving for velocity:
Since the athlete starts from rest, the initial velocity (v₀) is zero. The final velocity (v) when she reaches the platform can be found by substituting the time interval (0.8 s) into the equation:
v = v₀ + ∫F dt
v = 0 + [(9,200 N/s) * 0.5(0.8)^2 - (11,500 N/s^2) * (1/3)(0.8)^3]
Simplifying the expression:
v = (9,200 N/s) * 0.32 - (11,500 N/s^2) * (1/3) * 0.512
v ≈ 2,944 N - 615 N
v ≈ 2,329 N
Therefore, the athlete's velocity when she reaches the platform is approximately 2,329 N/s.
(b) To find the impulse the athlete received from the platform, we need to calculate the integral of the force function over the time interval she is in contact with the platform.
Impulse = ∫F dt
Impulse = ∫[(9,200 N/s)t - (11,500 N/s^2)t^2] dt (from t = 0 to t = 0.8s)
Evaluating this integral over the specified time interval will give us the impulse.
(c) To find the impulse the athlete received from gravity while in contact with the platform, we need to consider the impulse due to the change in momentum caused by gravity.
Impulse due to gravity = (change in momentum due to gravity)
Impulse due to gravity = mg * ∆t
Where m is the mass of the athlete (60.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and ∆t is the time interval the athlete is in contact with the platform (0.8 s).
(d) To find the velocity with which the athlete leaves the platform, we need to calculate the final velocity as she leaves the platform.
Using the equation v = v₀ + at, where v₀ is the final velocity on the platform (calculated in part a), a is the acceleration (which we can find using the force function), and t is the time interval she is in contact with the platform (0.8 s).
(e) To find the height to which she jumped upon leaving the platform, we need to use the equations of motion and consider the difference in potential energy before and after the jump.
The potential energy before and after the jump is given by PE = mgh, where m is the mass of the athlete, g is the acceleration due to gravity, and h is the height.
The difference in potential energy is equal to the work done by the force applied during the jump, which is the impulse received from the platform. Therefore, we can set the impulse equal to the difference in potential energy to solve for the height h.