A 70 kg man just before contact with the ground has a speed of 5 m/s In a stiff-legged landing he comes to a halt in 7ms (miliseconds). Keeping in mind that the average net force acting on the man includes both the force of the ground on the man and the gravity force, find the force of the ground on the man.
To find the force of the ground on the man, we need to use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Given:
Mass of the man (m) = 70 kg
Initial speed of the man (u) = 5 m/s
Final speed of the man (v) = 0 m/s (since he comes to a halt)
Time taken (t) = 7 ms = 7 × 10^(-3) s
First, we need to calculate the acceleration (a) of the man using the formula:
a = (v - u) / t
Substituting the given values, we have:
a = (0 - 5) / 7 × 10^(-3)
a = -5 / 7 × 10^(-3)
a = -714.29 m/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity. In this case, it means the man is decelerating.
Now, we can find the force acting on the man by multiplying his mass (m) by the acceleration (a):
Force = mass × acceleration
Force = 70 kg × (-714.29 m/s^2)
Force = -50,000 N
Therefore, the force of the ground on the man is -50,000 N. The negative sign indicates that the force is in the opposite direction to the initial motion of the man (resisting his motion).