A rope has mass density sigma (sigma= M/L, where M is the mass of the rope and L

Question

A rope has mass density sigma (sigma= M/L, where M is the mass of the rope and L is the length). The rope lies in a heap on the floor. You grab one end and pull straight up at constant speed v. a) What force must you apply, as a function of the height y that your end is above the floor? (Assume no knots or kinks, the rope comes straight up when you pull.) b) Does your answer make sense? How does it behave when y is very small and then increases to be large? You should get two terms in your answer: what, physically, does each one represent? Say something intelligent.

Explanation / Answer

When I pulling up thep rope

Than rate at which length of rope is moving up

r = dy/dt = V

Thus in every time interval extra momentum provided to the rope is

dP = v dm

dm =Mdy/L = Mvdt/L

So

dP = Mv2 dt /L

Thus

Extra Force required to provide this momentum

F = dP/dt = Mv2/L

So net force must be

Fnet = Myg /L + F

Or

Fnet =M /L(gy +v2) .........(answer)

If y is very very small than

Fnet = Mv2/L

The term of force containing Yg represent weight of the part of rope above ground and the term F represent impulsive force given to rope

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