What are the Work and Energies of Mass A and B in a Pulled System?

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SUMMARY

The discussion centers on the work-energy theorem as applied to two objects of different masses, A and B, being pulled by the same net force over the same distance. The work done on object A is established as 500 J, leading to the conclusion that the work done on object B is also 500 J, due to the equal force and distance. The kinetic energy of object A can be derived directly from the work done, while the kinetic energy of object B is also equal to the work done on it, affirming the principle that the change in work equals the change in kinetic energy.

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  • Basic knowledge of kinetic energy calculations
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  • Ability to manipulate algebraic equations
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Eternal Sky
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Homework Statement



Problem: Two objects of different mass start from rest, are pulled by the same magnitude net force, and are moved through the same distance. The work done on object A is 500 J. After the force has pulled each object, object A moves twice as fast as object B. (a) How much work is done on object B? (b) What is the kinetic energy of object A after being pulled? (c) What is the kinetic energy of object B after being pulled?

Homework Equations



[itex]W_n_e_t = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/itex]

The Attempt at a Solution



I believe that the answer to part (a) is 500 J also, since both objects are pulled through the same distance by the same force. I attempted to solve part (b) by using the above equation, but since I don't know the mass or the velocity, it didn't work out.

If someone could help me, I would greatly appreciate it.
 
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Eternal Sky said:
[itex]W_n_e_t = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2[/itex]

I believe that the answer to part (a) is 500 J also, since both objects are pulled through the same distance by the same force. I attempted to solve part (b) by using the above equation, but since I don't know the mass or the velocity, it didn't work out.

Hi Eternal Sky! :smile:

ah … the whole point of the work-energy theorem … ∆W = ∆KE … is that you don't need to know how to calculate KE …

(so you don't even need the whole of that equation of yours :wink:)

you just calculate ∆W, and ∆KE is automatically the same! :biggrin:
 
Ah, I see. Thanks a lot for your help!
 

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