Question

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 63.0 kg athlete jumps down onto the platform from a height of 0.690 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?

Answer 1

The force F by Newton's Second Law is F = dP/dt, or The net force on an object is the time rate of change of the momentum of the object.

Over the eight tenths of a second she is on the platform,

F = dP/dt = 9200N/s t - 11500 N/s^2 t^2
so

dP = Fdt = 9200N/s^2 t - 11500n/s^2 t^2

Integrating wrt t on the boundaries of 0 s to 0.8s will give the impulse she gives the table (N3 says this is equal in magnitude and opposite in direction to the impulse the table gives her)

integrating, we get

J = [4600N/s*t^2 - 3833N/s^2 * t^3] on boundaries 0 to 0.8s, so the impulse is (in magnitude), by evaluating the function at tf= 0.8s and ti = 0 we get the answer to (a)

(a) J = 981 Ns

The direction of the impulse the platform gives her is obviously upward because it has changed her velocity from downward to upward.

Now for (b): To get her speed as she reaches the platform, just use your equations of motion in the y direction. She falls from a height of 0.690 m, so her speed after this (starting from rest) is

V^2 = Vo^2 + 2*g*(yo-y), where g = 9.8m/s^2, Vo = 0, and (yo-y) = 0.690 m

so
V^2 = 2*9.8*0.690 m^2/s^2 = 13.5 m^2/s^2

and thus

(b) V = 3.68 m/s

Now for (c). Her impulse given to her by the platform changes her momentum. Initially her momentum is downward and is

Po = 63.0kg * V = 63.0kg * 3.68 m/s = 231.84 kgm/s = 231.84 Ns

so letting positive be upward her initial momentum is

Po = -231.84 Ns

Her final momentum after she leaves the table is given by

Pf - Po = J

so

Pf = Po + J = -231.84 Ns + 981 Ns = 749.16 Ns

So her speed upward when she leaves the platform is

(c) Vf = Pf/m = 749.16Ns/63.0kg = 11.89 m/s


(d) She will attain a height with that initial speed again given by

V^2 = Vf^2 - 2g(y-yo)

where V = 0 (the top of her jump)

and y-yo her height h

0 = Vf^2 - 2gh

2gh = Vf^2

h = Vf^2/2g = (11.89m/s)^2/(2*9.8m/s^2)

(d) h = 7.2 m