Answer:
[tex]3.9m/s^{2}[/tex]
Explanation:
Using second law of motion
[tex]a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2}[/tex] where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as [tex]9.81 m/s^{2}[/tex], T is torque and r is radius
Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain
[tex]a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}[/tex]
