Question

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

Answer 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?

Answer 2

Yes please

Answer 3

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 \(v_1\)

Answer 4

Why set them equal to each other

Answer 5

@Shadowys

Answer 6

It's because you do twice as much work.

Answer 7

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

Answer 8

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