Velocity of a 1-kg block after it has dropped 0.54 m

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Homework Help Overview

The discussion revolves around the dynamics of a coupled system involving two blocks of different masses, specifically focusing on the gravitational potential energy and kinetic energy as one block drops a certain distance. The original poster is attempting to calculate the velocity of a 1-kg block after it has dropped 0.54 m, while considering the effects of the coupled system on energy calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the calculation of gravitational potential energy using the equation k=mgh and question the validity of calculating kinetic energies independently for the coupled blocks. There is uncertainty about the direction of motion and the implications of the blocks being linked.

Discussion Status

Participants are actively questioning assumptions about the motion of the blocks and the nature of their coupling. Some guidance has been provided regarding the need to consider the system as a whole rather than treating the blocks separately. There is an ongoing exploration of how gravitational potential energy changes for each block and how that affects their kinetic energy.

Contextual Notes

There is confusion regarding the direction of motion and the impact of the blocks being coupled. Participants are also grappling with the implications of using gravitational acceleration in their calculations, especially in the context of one block moving upwards while the other moves downwards.

  • #31
vela said:
Faster than what?
Bad wording on my part, since kinetic energy is defined as 1/2 mv^2, wouldn't the 3.5kg block receive more of the system's kinetic energy since it has more mass, and both are moving at the same velocity?
 
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  • #32
volcore said:
Bad wording on my part, since kinetic energy is defined as 1/2 mv^2, wouldn't the 3.5kg block receive more of the system's kinetic energy since it has more mass, and both are moving at the same velocity?

Yes, exactly.
 
  • #33
PeroK said:
Yes, exactly.
So where do I go from there?

Would this be right:
(2.5kg)g(0.54m) = 1/2 m1v1^2 + 1/2m2v2^2
1.35g = 3.75v1^2 + 0.5v2^2?
 
  • #34
volcore said:
So where do I go from there?

Would this be right:
(2.5kg)g(0.54m) = 1/2 m1v1^2 + 1/2m2v2^2
1.35g = 3.75v1^2 + 0.5v2^2?

What do you know about the relationship between ##v_1## and ##v_2##?

Note that you should keep the units in these equations.
 

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