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3psilon Group Title

You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object’s speed. If you do work W, suppose the object’s final speed is v. What will be the object’s final speed if you do twice as much work? 1. Still v 2.√2 v 3. 4 v 4.v/√2 5. 2 v

  • 2 years ago
  • 2 years ago

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  1. Shadowys Group Title
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    notice that \(W=\Sigma E_k\) so \(W=\frac{1}{2}m(v^2 -u^2)\) since u=0,\(W=\frac{1}{2}m(v^2)\) Do you need further help?

    • 2 years ago
  2. 3psilon Group Title
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    Yes please

    • 2 years ago
  3. Shadowys Group Title
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    okay, \(W_1=\frac{1}{2}m(v_1^2)\) \(W_2=\frac{1}{2}m(v_2^2)\) now, \(W_2=2W_1\) \(2W_1=\frac{1}{2}m(v_2^2)\) \(m(v_1^2)=\frac{1}{2}m(v_2^2)\) find \(v_2\) in terms of \(v_1\)

    • 2 years ago
  4. 3psilon Group Title
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    Why set them equal to each other

    • 2 years ago
  5. 3psilon Group Title
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    @Shadowys

    • 2 years ago
  6. Shadowys Group Title
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    It's because you do twice as much work.

    • 2 years ago
  7. 3psilon Group Title
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    how are the works equal if you're doubling the work

    • 2 years ago
  8. Shadowys Group Title
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    the second work is equal to twice the original work. Putting that in terms of algebra is \(W_2 =2W_1\)

    • 2 years ago
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