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3psilon

  • 2 years ago

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

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  1. Shadowys
    • 2 years ago
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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. 3psilon
    • 2 years ago
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    Yes please

  3. Shadowys
    • 2 years ago
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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 \(v_1\)

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

  5. 3psilon
    • 2 years ago
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    @Shadowys

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

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

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

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