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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
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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Shadowys Group TitleBest 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?
 2 years ago

Shadowys Group TitleBest 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 \(v_1\)
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Why set them equal to each other
 2 years ago

Shadowys Group TitleBest ResponseYou've already chosen the best response.1
It's because you do twice as much work.
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
how are the works equal if you're doubling the work
 2 years ago

Shadowys Group TitleBest 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 \(W_2 =2W_1\)
 2 years ago
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