What Is the Person's Velocity After the First 25 cm of the Chin-Up?

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The discussion focuses on calculating the velocity of a person performing a chin-up after lifting their body 25 cm. The individual weighs 700N, and each arm exerts an upward force of 355N, resulting in a net force of 10N. The work done during the lift is calculated as 2.5 J, leading to a velocity of 0.035 m/s. Participants suggest using the work-energy principle (W = ΔK) and kinematic equations to further analyze the motion.

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Original Question: A person doing a chin-up weighs 700N exclusive of the arms. During the first 25.0 cm of the lift, each arm exerts an upward force of 355N on the torso. If the upward movement starts from rest, what is the persons velocity at the point.

some work:
700-2(355) = 10 N
W = 25 cm *10 N
.25M*10N= 2.5 J
(Mass ~ 700N = 9.8[M] )
A = 2.5 / 71.4
=.035 m/s

My question: so what do I do now with this .035 m/s?
 
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Try using [tex]W = \Delta K[/tex]
 
You could also find acceleration using [itex]F_{net}=ma[/itex] then use kinematic formulas to find final velocity.
 

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