Question

A 66 kg human sprinter can accelerate from rest to 11 m/s in 3.4s. During the same interval, a 35 kg greyhound can accelerate from rest to 20 m/s. Compute

(a) the change in kinetic energy of the sprinter.
(b) the change in kinetic energy of the greyhound.
(c) the average power output of the sprinter.
(d) the average power output of the greyhound.

Answer 1

Human Sprinter:
a = (V-Vo)/t = (11-0)/3.4 = 3.24 m/s^2.
d = 0.5a*t^2 = 1.62*(3.40)^2 = 16.96 m.
Fs = m*g = 66kg * 9.8N/kg = 646.8 N. =
Force of sprinter.

Greyhound Sprinter:
a = (20-0)/3.4 = 5.88 m/s^2.
d = 2.94*(3.4)^2 = 34 m.
Fg = 35kg * 9.8N/kg = 343 N. = Force of
Greyhound.

a. KE Change=0.5m*V^2-0.=33*11^2=3993 Joules.

b. KE Change=17.5*20^2=70,000 Joules.

c. Po = Fs*d/t.
Po = 646.8 * 16.96/3.4 = 3226 J/s.

d. Po = Fg*d/t.
Po = 343 * 34/3.4 = 3430 J/s.

Answer 2

Change in Kinetic Energy ΔK = Kf - Ki
Kf = 1/2 * m * vf^2
Ki = 1/2 * m * vi^2 (vi is zero, so Ki is zero)
Therefore ΔK = Kf - 0 = 1/2 * m * vf^2

Average power over a time interval Δt is ΔE/Δt .
ΔE = ΔK
Therefore Average Power = ΔK/Δt