(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A uniform rope of mass λ per unit length is coiled on a smooth horizontal table. One end is pulled straight up with constant speedv_{0}

Find the force exerted on the end of the rope as a function of heightyand find the power delivered to the rope.

2. Relevant equations

F_{net}= ma

F_{gravity}= mg

ΔK = ∫F_{net}dy

P = Fv

3. 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 = λyg

P = 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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# Homework Help: Force/Power pulling a rope (I get 2 solutions)

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