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 function

F = 9 200t − 11 500t2

where 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

a).
`I=981J`

b).
`V_f=3.756m/s`

c).
`v_1=3.04m/s`

d).
`y'=0.47m`

Step-by-step explanation:

Given:

`m=67kg`,
`h_i=0.720m`,
`t_1=0s`,
`t_2=0.8s`

a).

`F=9200t-11500t^2`

`I=\int\limits^a_b {F} \, dx =\int\limits^.8_0 {9200t-11500t^2} \, dt`

`I=4600t^2-3833.33t^3|0,0.8`

`I=981J`

b).

`v_f^2=v_i^2+2*a*y`

She began in rest so vi=0

`v_f=√(2*g*y)=√(2*9.8m/s^2*0.720m)=√(14.112 m^2/s^2)`

`V_f=3.756m/s`

c).

Impulse total=momentum total

`I_i-I_f=m_1*v_1-m_1*v_f`

`981-(67kg*9.8m/s^2*0.8)=67kg*v_1-67*kg*(-3.756m/s)`

Solve to v1

`v_1=(204.068kg*m/s)/(67kg)`

`v_1=3.04m/s`

d).

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

`v_f=0`

`y=(-v_i^2)/(2*g)=(-(3.04m/s)^2)/(2*-9.8m/s^2)`

`y'=0.47m`