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 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
 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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Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1notice 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?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1okay, \(W_1=\frac{1}{2}m(v_1^2)\) \(W_2=\frac{1}{2}m(v_2^2)\) 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 \(v_1\)

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

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

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

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