A soccer player kicks a soccer ball (mass 0.40 kg, initially at rest) in the hor

Question

A soccer player kicks a soccer ball (mass 0.40 kg, initially at rest) in the horizontal (+x) direction. A small force sensor inside the ball measures the force applied to the ball vs. time. You perform a curve fit on this data and find the following function is an excellent approximation for the forward (positive) force applied by the player's foot.
Units are omitted; assume that input of [t]=s yields [F]=N. Recommended: do a rough sketch of this graph for reference.

Fx(t) = (4 · 10^6) t - (1 · 10^9) t^2 for Fx = 0
Fx(t) = 0 otherwise.

(a) Identify the time region of interest:
At what time does the kick start? ms
At what time does the kick end? ms


(b) During the kick, what is the maximum force exerted by the player's foot on the ball? Also, find the time it occurs.
Fmax = N
occurs at t = ms


(c) Find the impulse imparted to the ball during this kick.
?px = N·s

(d) During the kick, what is the average force exerted by the player's foot on the ball?
Fav = N


(e) Find the ball's velocity immediately after it leaves the player's foot.
vf,x = m/s

Explanation / Answer

at the time of kick starting and kick ending force should be 0 solvind force=o give t=0 and t=0.004 sec so kick started at t=0 kick ended at t=0.004 sec or 4ms d(F)/dt=(4 · 10^6) - 2*(1 · 10^9) t put d(F)/dt=0 to get the max force t=2ms Fmax=8*10^3-4*10^3=4000N to calculate impulse calculate integral of Fdt from t=0 to 4 ms Impulse(t)=(2*10^6)t^2-(1*10^9)*t^3/3 limits from t=0 to t=4 ms Impulse total=(2*10^6)*(16*10^(-6))-(10^9)*(64*10^(-9))/3 =32-64/3= 10.67 Ns avg force=impulse total/total time=10.67/.004=2667.5 N impulse= change in momentum 10.67=mv=0.40*v v=26.675 m/sec

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