A runner whose mass is 49 kg accelerates from a stop to a speed of 6 m/s in 3 seconds. (A good sprinter can run 100 meters in about 10 seconds, with an average speed of 10 m/s.)

Question

(a) What is the average horizontal component of the force that the ground exerts on the runner's shoes?

(b) How much displacement is there of the force that acts on the sole of the runner's shoes, assuming that there is no slipping?

c) How much work is done on the point-particle system by this force?

Answer 1

force * time = change of momentum

a) force = 49 * (6-0) / 3
= 98 N

b) force acts on the average over one hundred meters

c) work = force * distance = 98 * 100 = 9800 Joules

interesting, lets see how much kinetic energy we have
(1/2) m v^2 = .5 * 49 * 36 = 882 Joules
s we are not very efficient with our fluctuating force.

Answer 2

(a): Force = Mass * Acceleration
Acceleration = Velocity / time
Force = ?
Mass = 49kg
Velocity = 6m/s
Time = 3sec
Distance = 100m
Acceleration = 6/3 = 2m/s^2
Force = (49kg) * (2m/s^2) = 98N

(b)

(c) W = Force * distance
W = Ek
Ek = 1/2 * mass * velocity^2
Ek = 1/2 * 49 * 6 * 6 = 882 Joules

# The Chemical Energy Decreases as the Kinetic Energy of the Runner Increases