A5.33 kg mass is attached to a string which is wound around a fixed
pulley (e. circular disc) which has a mass of 233 kg and a radius of 0486
m, as seen in the figure above. the mass is released and falls, causing the
pulley to rotate, solve for (a) the acceleration of the block,

0.429 m/s²

Explanation:

Draw a free body diagram of the pulley and the block.  The pulley has one force acting on it (tension downward).  The block has two forces acting on it (tension upward and weight downward).

Sum of the torques on the pulley:

∑τ = Iα

Tr = (½ Mr²) (a / r)

T = ½ Ma

Sum of the forces on the block (in the downward direction):

∑F = ma

mg − T = ma

Substituting:

mg − ½ Ma = ma

mg = ½ Ma + ma

mg = (½ M + m) a

a = mg / (½ M + m)

Given g = 9.8 m/s², M = 233 kg, and m = 5.33 kg:

a = (5.33 kg) (9.8 m/s²) / (½ (233 kg) + 5.33 kg)

a = 0.429 m/s²

v 154.9= 0.9 length length s little over 154.9cm

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this