Kleppner question on Power

  • Thread starter RubinLicht
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Homework Statement


This is actually two problems, both of which i have already successfully solved, I just have conceptual questions about the results.

5.18 Sand and conveyor belt Sand runs from a hopper at constant rate dm/dt onto a horizontal conveyor belt driven at constant speed V by a motor.
(a) Find the power needed to drive the belt.
(b) Compare the answer to (a) with the rate of change of kinetic energy of the sand. Can you account for the difference?
Answer: P = 2 dK/dt oh wow this color is nice on the eyes

5.19 Coil of rope A uniform rope of mass density λ per unit length is coiled on a smooth horizontal table. One end is pulled straight up with constant speed v0, as shown.
(a) Find the force exerted on the end of the rope as a function of height y.
(b) Compare the power delivered to the rope with the rate of change of the rope’s total mechanical energy.
Answer: P= dE/dt + 1/2λVo3
dE/dt = 1/2λVo3
+ygvo

Homework Equations


P=Fv
F=dP/dt

The Attempt at a Solution


What i can think of is that the power is larger than the net change in energy because non elastic collisions occur between the belt and sand (and rope "links", if you think about a rope as a chain of infinitesimal rings) during the jerky period of acceleration. Please correct me if I am wrong. I encountered the conveyor belt problem as an example in Resnick and Halliday, but they did not explain either. Although i have a hazy idea of whats going on, I would very much appreciate it if someone cleared up the confusion I've been feeling about these problems. Thank you. (collisions are in the next chapter, so I haven't formally covered them yet)
 
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Answers and Replies

  • #2
TSny
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5.18 Sand and conveyor belt Sand runs from a hopper at constant rate dm/dt onto a horizontal conveyor belt driven at constant speed V by a motor.
(a) Find the power needed to drive the belt.
(b) Compare the answer to (a) with the rate of change of kinetic energy of the sand. Can you account for the difference?
Answer: P = 1/2 dK/dt
This says that the power is less than the rate of change of K. Is that correct?
oh wow this color is nice on the eyes
:cool:

The Attempt at a Solution


What i can think of is that the power is larger than the net change in energy because non elastic collisions occur between the belt and sand (and rope "links", if you think about a rope as a chain of infinitesimal rings) during the jerky period of acceleration.
Yes, that's exactly right. That's good insight for not having covered collisions yet! The power is the rate at which total energy of the system changes. But the total energy includes more than just mechanical forms of energy.
 
  • #3
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Woops typo, thanks for the reply.
 
  • #4
TSny
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Woops typo
Yes, I knew that. I was just "ribbing" you a little.
 
  • #5
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Yes, I knew that. I was just "ribbing" you a little.
Heh, thanks for the help.
 

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