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 67.0-kg athlete jumps down onto the platform from a height of 0.720 m. While she is in contact with the platform during the time interval 0 < t < 0.800 s, the force she exerts on it is described by the functionF = 9 200t − 11 500t2where F is in newtons and t is in seconds.(a) What impulse did the athlete receive from the platform?N · s up(b) With what speed did she reach the platform?m/s(c) With what speed did she leave it?(d) To what height did she jump upon leaving the platform?

Answer 1

Answer:

a.$I=981.34 N*s$

b.$v_f=3.96 m/s$

c.$v_{f1}=3.63m/s$

d.$y_f=0.673m$

Explanation:

Given: $m=67kg$, $h=0.720m$, $0$

a.

$I=\int\limits^{t_1}_{t_2} {F(t)} \, dt$

$F(t)=9200*t-11500t^2$

$I=\int\limits^{0.8s}_{0s}{9200*t-11500*t^2} \, dt$

$I=4600*t^2-3833.3*t^3|(0.80,0)$

$I=2944-1962.66=981.35$

$I=981.34 N*s$

b.

$v_f^2=v_i^2+a*y'$

Starting from the rest

$v_f^2=0+2*9.8m/s^2*0.80s$

$v_f^2=15.68$

$v_f=\sqrt{15.68m^2/s^2}=3.96 m/s$

c.

$I_{total}=p_f$

$I_1-m*g*d=m*v_{f1}-m*v_f$

$981.34-67kg*9.8m/s^2*0.720=67.0kg*v_{f1}-67.0kg*(-3.96m/s)$

Solve to vf

$v_{f1}=3.63m/s$

d.

$v_f^2=v_i^2+2*a*y_f'$

$y_f'=v_i/2*a =(3.63m/s)^2/2*9.8m/s^2$

$y_f=0.673m$