Physics: If a 65-kg man undergoes a turning acceleration of 5 m/s2 during a running turn, what is the magnitude of force experienced by the foot due to the ground?

If a 65-kg man undergoes a turning acceleration

Chynnab7416

05.01.2022

Force, F = 715.11 newton

Explanation:

Given that,

Mass of the person, m = 65 kg

Acceleration of man during a running turn, $a=5\ m/s^2$

We need to find the magnitude of force experienced by the foot due to the ground. It can be calculated using second law of motion as :

F = ma

$F=65\ kg\times 5\ m/s^2$

F = 325 Newton

Here, the weight of the person and this force force a right angle triangle.

The wight of the person, W = mg

$W=65\ kg\times 9.8\ m/s^2=637\ N$

So, the resultant for is, $F_r=\sqrt{325^2+637^2}$

$F_r=715.11\ N$

So, the f force experienced by the foot due to the ground is 715.11 newtons. Hence, this is the required solution.

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