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

1350N

Step-by-step 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 = (\Delta P)/(\Delta t) = (m(0 - v))/(\Delta t) = (1.65*(2.25 - 0))/(2.75 * 10^(-3)) = -1350 N`

The magnitude of the force would be 1350N