Answer:
[tex]6.5\; {\rm N}[/tex].
Explanation:
When an object travel at a speed of [tex]v[/tex] in a circle of radius [tex]r[/tex], the (centripetal) acceleration of that object would be [tex]a = (v^{2} / r)[/tex].
In this question, the ball is travelling at [tex]v = 3\; {\rm m\cdot s^{-1}}[/tex] in a circle of radius [tex]r = 1.8\; {\rm m}[/tex]. The (centripetal) acceleration of this ball would be:
[tex]\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
By Newton's Laws of Motion, for an object of mass [tex]m[/tex], if the acceleration of that object is [tex]a[/tex], the net force on that object would be [tex]m\, a[/tex]. Since the acceleration of this ball is [tex]a = 5\; {\rm m\cdot s^{-2}}[/tex], the net force on this ball would be:
[tex]\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}[/tex].
