Question

A person doing a chin-up weighs 660 N, exclusive of the arms. During the first 22.0 cm of the lift, each arm exerts an upward force of 355 N on the torso. If the upward movement starts from rest, what is the person's velocity at this point

Answer 1

0.57183 m/s

Step-by-step explanation:

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration

u = Initial velocity

v = Final velocity

s = Displacement

Mass of person

`m=(W)/(g)\\\Rightarrow m=(660)/(9.81)\\\Rightarrow m=67.27828\ kg`

As the forces are conserved

`\text{Upward forces - Downward forces}=ma\\\Rightarrow \text{Force of both arms - Force of gravity}=ma\\\Rightarrow 355+355-660=67.27828a\\\Rightarrow 50=67.27828a\\\Rightarrow a=(50)/(67.27828)\\\Rightarrow a=0.74318\ m/s^2`

From equation of motion

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* 0.74318* 0.22+0^2)\\\Rightarrow v=0.57183\ m/s`

The person's velocity at the point is 0.57183 m/s