3psilon
  • 3psilon
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
Physics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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?
3psilon
  • 3psilon
Yes please
anonymous
  • anonymous
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 \(v_1\)

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3psilon
  • 3psilon
Why set them equal to each other
3psilon
  • 3psilon
@Shadowys
anonymous
  • anonymous
It's because you do twice as much work.
3psilon
  • 3psilon
how are the works equal if you're doubling the work
anonymous
  • anonymous
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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