Verified answer

Let moment of inertia of the drum about its horizontal axis be I.

M= mass of the block,

a = acceleration of the block = a= 3.7 m/s^2

T = tension in the rope

Considering the forces acting on the block,

Mg-T = Ma (T in the rope always acts in a direction away from the block)

T = M (g-a) = 50 * (9.8-3.7) N = 305 N

Angular Acceleration of the drum = linear acceleration / radius of drum = a/r

Torque acting on the drum = T*r = angular acceleration X Moment of inertia

T*r = (a/r) * Mr^2

M = T/a = 305/3.7 kg = 82.4 Kg

Hence both your problems are solve. Here I have assumed that rope is not slipping over the drum.

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