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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 one year ago one year ago

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1. Shadowys Group Title

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. 3psilon Group Title

Yes please

3. Shadowys Group Title

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 $$v_1$$

4. 3psilon Group Title

Why set them equal to each other

5. 3psilon Group Title

@Shadowys

6. Shadowys Group Title

It's because you do twice as much work.

7. 3psilon Group Title

how are the works equal if you're doubling the work

8. Shadowys Group Title

the second work is equal to twice the original work. Putting that in terms of algebra is $$W_2 =2W_1$$