maryam geraminejad asks:

CATAGORY: Mechanics QUESTION: Lets say a guy is hanging still from a single vertical rope. He starts to climb upwards. I understand that the tension upwards is initially equal to the downward force mg so that the guy is stationary. But how is it that he is able to pull himself upwards with a velocity that he initiated into the system without there being a net positive force upwards? I can calculate the problems by using newtons laws and plugging stuff in, but conceptually it seems strange to me. Please help. s=distance The confusion started when I had to work several power problems of this climbing rope type where P=W/T= ma(s)/T where we must assume a=9.8

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First of all lets clarify the equation you wrote.

(1) P=W/T= ma(s)/T

P is the power which has units of energy per unit time so the T part
is the time that it takes to change the gravitational potential energy
by ma(s) where m is the mass of the lifted object, a = 9.8m/s^2 is the
magnitude of the acceleration due to gravity and s is the distance
through which the object is lifted.  ma is is the NET force on the
mass.  So long as the person is climbing with CONSTANT velocity the
downward force ma is the only force acting on the persons mass so the
tension in the rope is always ma.  The initial part where the person
accelerates up to speed however is a different matter.  If a spring
scale were attached to the top of rope the scale would read a value
higher than ma during the acceleration.  What most texts assume
however is that this distance over which this acceleration occours is
so short that we can approximate the power with formula (1) above.

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