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

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In the discussion about the work and energies of two objects pulled by the same force, it is established that the work done on object A is 500 J. Since both objects are subjected to the same force over the same distance, the work done on object B is also 500 J. The kinetic energy of object A can be determined directly from the work done, as the work-energy theorem states that the change in work equals the change in kinetic energy. The participants clarify that specific mass or velocity values are unnecessary for calculating kinetic energy in this scenario. The conversation emphasizes the application of the work-energy theorem in solving the problem.
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



W_n_e_t = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

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:
W_n_e_t = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

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!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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