Two blocks with different mass are attached to either end of a light rope that passes

  • Thread starter kdizzle711
  • Start date
  • #1
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Homework Statement


Two blocks with different mass are attached to either end of a light rope that passes over a light, frictionless pulley that is suspended from the ceiling. The masses are released from rest, and the more massive one starts to descend. After this block has descended a distance 1.30 m, its speed is 3.50m/s .

If the total mass of the two blocks is 14.0 kg, what is the mass of the more massive block?



Homework Equations



(1/2)mv1^2+mgy1=(1/2)mv2^2+mgy2

The Attempt at a Solution


Can someone help me get started with this problem? I'm not sure I am using the right equation or approaching it correctly
 

Answers and Replies

  • #2
279
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Since you know that a 14 Kg mass has been accelerated from rest to 3.50 m/s over a distance of 1.30 m, you can detemine the acceleration from the equation:
[tex]v^{2} = u^{2} + 2as[/tex]

From there you can calculate the force required to accelerate the mass at that rate with:
[tex]F = ma[/tex]

That will give you the difference in the weights of the two masses.
 
  • #3
27
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What does u^2 stand for?
 
  • #4
279
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v is the final velocity, u is the initial velocity. Perhaps you use a different type of notation. The same formula is listed last here in a different notation.
 
  • #5
27
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I found that the F=ma is 65.94N, but where do I go from there?
 
  • #6
279
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Like I said, that's the difference in their weights (though I actually get 65.96 N). You can divide that by g to find the difference in their mass since weight = mass * gravitational field strength.

If you call the two weights a and b, that will give you:
a - b = 65.96/g

You also have
a + b = 14

Now you have to solve the simultaneous equations.
 
  • #7
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Thanks, I got it. You guys are amazing
 
  • #8
279
1
Nah, not amazing just ... yeah OK, amazing. :wink:

Happy to help. :smile:
 
  • #9
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This appears to be a common question. I got the same thing with different values, but the formulae here worked!
 

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