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 2.25 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 1.65 kg ?

Answer 1

Answer:

1350N

Explanation:

$2.75 ms = 2.75*10^{-3}s$

The force exerted on the hand would be the momentum divided by the duration of contact.

As the hand is coming to rest, final velocity would be 0

$F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N$

The magnitude of the force would be 1350N