Question

A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms from an initial speed of 3.75 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.75 kg ?

Answer 1

Explanation:

Given that,

Initial speed of the man's hand, u = 3.75 m/s

Finally, the hand comes to rest, v = 0

Time interval, t = 2.75 ms

Total mass of the forearm and the hand is, m = 1.75 kg

To find,

The magnitude of force exerted on the leg

Solution,

We know that the force exerted from one object to another is given by using second law of motion as :

F = m × a

a is the acceleration of the object

$F= m\times \dfrac{v-u}{t}$

$F= 1.75\times \dfrac{0-3.75}{2.75\times 10^{-3}}$

F = -2386.36 N

Therefore, the force exerted on the leg is 2386.36 N.