Question

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 72.9-kg man just before contact with the ground has a speed of 8.74 m/s. (a) In a stiff-legged landing he comes to a halt in 1.71 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.268 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b)

Answer 1

mass =m= 72.9 Kg

a)

Stiff-legged landing

Impact time between leg and ground = = 0.171 milli sec = 0.171 x 10^(-3) secs

Speed before contact =u= 8.74 m/s

Speed at end =v= 0 m/s

Change of momentum = m(v-u)

Average force = m(v-u)/ = 72.9 Kg ( 8.74 m/s)/0.171 x 10^(-3) sec= 3726000 Newtons (upwards)

b)

Bent legged landing

Impact time between leg and ground = =0.268 seconds

Average force on legs = Rate of change of momentum = m(v-u)/ = 72.9 Kg x (8.74m/s)/ 0.268 s = 2377 Newtons (upward)

c)

Average force applied by ground on the man =N =?

Average net force on leg in case b = 2377 N

This net force net force on leg including impact force up wards and gravity downwards = 2377N = N - 72.9Kg x 9.8 m/s

F = 2377 N + (72.9Kg x 9.8 m/s^2) = 3091 N