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 70.8-kg man just before contact with the ground has a speed of 6.33 m/s. (a) In a stiff-legged landing he comes to a halt in 1.99 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 0.147 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

(a) The man comes to a halt in 1.99 ms = 1.99 × 10⁻³ s, so his average acceleration slowing him from 6.33 m/s to a rest is

a = (6.33 m/s - 0) / (1.99 × 10⁻³ s) ≈ 3180 m/s²

so that the average net force on the man during the landing is

F = (70.8 kg) a ≈ 225,000 N

i.e. with magnitude 225,000 N.

(b) With knees bent, the man has an average acceleration of

a = (6.33 m/s - 0) / (0.147 s) ≈ 43.1 m/s²

and hence an average net force of

F = (70.8 kg) a ≈ 3050 N

(c) The net force on the man is

∑ F = n - w = m a

where

n = magnitude of the normal force, i.e. the force of the ground pushing up on the man

w = the man's weight, m g ≈ 694 N

m = the man's mass, 70.8 kg

g = mag. of the acceleration due to gravity, 9.80 m/s²

a = the man's acceleration

Using the acceleration in part (b), we have

n = m g + m a = m (g + a)

n = (70.8 kg) (9.80 m/s² + 43.1 m/s²) ≈ 3740 N