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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
 one year ago
 one year 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
 one year ago
 one year ago

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ShadowysBest ResponseYou've already chosen the best response.1
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?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
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 \(v_1\)
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
Why set them equal to each other
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
It's because you do twice as much work.
 one year ago

3psilonBest ResponseYou've already chosen the best response.0
how are the works equal if you're doubling the work
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
the second work is equal to twice the original work. Putting that in terms of algebra is \(W_2 =2W_1\)
 one year ago
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