$\begin{array}{1 1} 21.7\;m/s \\ 9.3\;m/s \\ 2.7\;m/s \\3.6\;m/s \end{array} $

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Answer : 9.3 m/s

Step 1:

Two forces are acting on the car, gravity and the normal force.

At the top, both forces are toward the center of the circle, so Newtonâ€™s second law gives $mg+n_B=ma_{rad}$

At the bottom, gravity is downward but the normal force is upward, so $n_A-mg=ma_{rad}$

Solving for the acceleration

$a_{rad} =g+ \large\frac{n_B}{m}$

$\qquad= 9.8 m/s +\large\frac{6N}{0.8 \;kg}$

$\qquad= 17.3 \;m/s^2$

Step 2:

The speed of the car can now be calculated :

$v= \sqrt {a_{rad} R}$

$\quad= \sqrt {17.3 m/s^2.5 \;m}$

$\qquad= 9.3 m/s$

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