- #1
Ginny Mac
- 17
- 0
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?
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
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