A force platform is a tool used to analyze the performance of athletes measuring

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 68.0-kg athlete jumps down onto the platform from a height of 0.550 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? m/s

(d) To what height did she jump upon leaving the platform? m

Explanation / Answer

a) impulse = integral of F.dt

        = integral of ( 9200t - 11500t^2) dt

     = 4600t^2 - 3833.33t^3

t is from 0 to 0.8 s

impulse=4600(0.8^2) - 3833.33(0.8^3) = 981.34 N s

b) using energy conservation to find reach speed.

mgh = mv^2 /2

9.81 x 0.550 = v^2 / 2

v = 3.28 m/s

c) impulse = change in momentum

981.34 = 68( vf - ( - 3.28))

vf = 11.15 m/s

d) using energy conservation to find height,

mgh = mv^2 /2

9.81 x h = 11.15^2 / 2

h = 6.34 m

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