A chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length L= 28 cm and mass m=0.012 kg, how much work is required to pull the hanging part back onto the table?(adsbygoogle = window.adsbygoogle || []).push({});

I have used this model: W horizontal + Work due to gravity = Work

So...

Wg = (0.012 kg)(9.8 m/s^2)(.07 m)(cos 180)

= -5.0 * 10^-3 J

Then I used the work-kinetic energy theorem (change in K = W) because we do not know anything about the force applied to the chain. I get:

change in K = W due to the horizontal force + Work due to grav. + Work due to normal force

change in K is zero, because the object is stationary before and after the pull, so...

0= (W due to horizon. force) - (5.0 * 10^-3) + (5.0 * 10^-3), so...

Work due to force = 0

I just get zero work for the horizon. force. Is the only work done here the Work due to gravity? Should I just leave it at -5.0 * 10^-3 J, if that answer is even correct?? Hmm....

Thank you,

Ginny

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# Homework Help: Work required to pull a hanging chain

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