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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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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?
Yes please
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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Other answers:

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