$\begin{array}{1 1} 12.2\;m/s \\ 3.5\;m/s \\ 4.8\;m/s\\ 2.9\;m/s\end{array} $

Answer : $ 12.2\;m/s $

The force causing the centripetal acceleration in this case is the force T exerted by the cord on the ball, Solving for v, we have

$T= m \large\frac{v^2}{r}$

$v=\sqrt { \large\frac{Tr}{m}}$

$v_{max} =\sqrt { \large\frac{T_{max}T}{m}}$

$\qquad= \sqrt { \large\frac{(50.0 \;N)(1.50\;m)}{0.500\;kg}}$

$\qquad= 12.2 \;m/s$

Ask Question

Tag:MathPhyChemBioOther

Take Test

...