Answer:
1. four times as much force as required to keep the slower object on the path.
Explanation:
The centripetal force that keeps the object on a circular path is given by
[tex]F=m\frac{v^2}{r}[/tex]
where
m is the mass of the object
v is its tangential speed
r is the radius of the circular path
In this problem, we have a second identical object (so, same mass) that moves around a circle of same diameter of the first one (so, same radius), but with a speed that is twice the speed of the first one: v' = 2v. Therefore, its centripetal force will be
[tex]F'=m\frac{v'^2}{r}=m\frac{(2v)^2}{r}=4m\frac{v^2}{r}=4F[/tex]
So, the centripetal force required to keep the second object on the circular path is
1. four times as much force as required to keep the slower object on the path.