Answer:
![√(2)v](https://img.qammunity.org/2020/formulas/physics/high-school/wgbbsfkvfx790l772gxlv6gbg84dzwbc4b.png)
Step-by-step explanation:
The work done on the object at rest is all converted into kinetic energy, so we can write
![W=(1)/(2)mv^2](https://img.qammunity.org/2020/formulas/physics/high-school/ukygyl53ltr8nbsj6ak83z6s5bvcs2wroc.png)
Or, re-arranging for v,
![v=\sqrt{(2W)/(m)}](https://img.qammunity.org/2020/formulas/physics/high-school/s6bkqijzxa2uw0fs3d3c4rjhrllbnep1pr.png)
where
v is the final speed of the object
W is the work done
m is the object's mass
If the work done on the object is doubled, we have W' = 2W. Substituting into the previous formula, we can find the new final speed of the object:
![v'=\sqrt{(2W')/(m)}=\sqrt{(2(2W))/(m)}=√(2)\sqrt{(2W)/(m)}=√(2)v](https://img.qammunity.org/2020/formulas/physics/high-school/ky6tjrwvm9619soe0uc564f3kj5se4c5v1.png)
So, the new speed of the object is
.