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## Homework Statement

A uniform rope of mass λ per unit length is coiled on a smooth horizontal table. One end is pulled straight up with constant speed

*v*

_{0}Find the force exerted on the end of the rope as a function of height

*y*and find the power delivered to the rope.

## Homework Equations

F

_{net}= ma

F

_{gravity}= mg

ΔK = ∫F

_{net}dy

P = Fv

## The Attempt at a Solution

__Solution 1:__"with constant speed v

_{0}" [itex]\Rightarrow[/itex] F

_{net}= 0 [itex]\Rightarrow[/itex]

**F**

_{pull}= F_{gravity}= m(y) * g = λygP = F

_{pull}v = λgyv

_{0}

__Solution 2:__ΔK = ∫F

_{net}dy

[itex]\frac{1}{2}[/itex]m(y) * v

_{0}

^{2}= ∫F

_{net}dy

[itex]\frac{1}{2}[/itex]m(y) * v

_{0}

^{2}= ∫(F

_{pull}- F

_{gravity})dy

[itex]\frac{1}{2}[/itex]λyv

_{0}

^{2}= ∫F

_{pull}dy - ∫F

_{gravity}dy

[itex]\frac{1}{2}[/itex]λyv

_{0}

^{2}= ∫F

_{pull}dy - m(y) * g

[itex]\frac{1}{2}[/itex]λyv

_{0}

^{2}= ∫F

_{pull}dy - [itex]\frac{1}{2}[/itex]λy

^{2}g (where the half comes from the fact that gravity acts at the center of mass of the rope (y/2)).

[itex]\frac{1}{2}[/itex]λyv

_{0}

^{2}+ [itex]\frac{1}{2}[/itex]λy

^{2}g = ∫F

_{pull}dy

Differentiate both sides with respect to y:

**[itex]\frac{1}{2}[/itex]λv**

_{0}^{2}+ λyg = F_{pull}P = F

_{pull}v = λv

_{0}

^{3}+ λygv

_{0}

So which ones right? Please tell me why its right and also why the other one is wrong.

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