Answer:
No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional to the speed squared, the speed will be
times larger.
Step-by-step explanation:
The work done on the ball is equal to the kinetic energy gained by the ball:
![W=K](https://img.qammunity.org/2020/formulas/physics/high-school/3e39jyfyxmkln0920lzfd9vm1oavrtwf0v.png)
So when the work done doubles, the kinetic energy doubles as well:
![2W = 2 K](https://img.qammunity.org/2020/formulas/physics/high-school/gpa4acf0jubandx5i49oew7zefb677meac.png)
However, the kinetic energy is given by
![K=(1)/(2)mv^2](https://img.qammunity.org/2020/formulas/physics/middle-school/c6fs3acuplloc3whu5cpc8ui63cnl7ur39.png)
where
m is the mass of the ball
v is its speed
We see that the kinetic energy is proportional to the square of the speed,
. We can rewrite the last equation as
![v=\sqrt{(2K)/(m)}](https://img.qammunity.org/2020/formulas/physics/high-school/gw6xy3xoxjuxpwgvnb78006o53ak8sjrub.png)
which also means
![v=\sqrt{(2W)/(m)}](https://img.qammunity.org/2020/formulas/physics/high-school/s6bkqijzxa2uw0fs3d3c4rjhrllbnep1pr.png)
If the work is doubled,
![W'=2W](https://img.qammunity.org/2020/formulas/physics/high-school/48h0twvgn5wfvjsz2vd0nfm47cif2bhi00.png)
So the new speed is
![v'=\sqrt{(2(2W))/(m)}=√(2)\sqrt{(2W)/(m)}=√(2) v](https://img.qammunity.org/2020/formulas/physics/high-school/rpwndpag7usxski44fc4074o7fup3dgq9p.png)
So, the speed is
times larger.