Question

A load of mass M1 = 2.3kg is attached to a mass M2=4.25 with a light inextensible string hanging over a frictionless pulley.If the two loads hang freely in air what is the acceleration of m1 upwards

Answer 1

We may conclude the smaller load to move upwards, so its acceleration $a$ is directed upwards. Let us now determine the forces that act on the loads.

M1: $M_1 a = T-M_1g$ , where T is the tension of string,

M2: $M_2a = M_2g-T,$ therefore,

$(M_1+M_2)a = (M_2-M_1)g, \;\; \\ a = \dfrac{M_2-M_1}{M_1+M_2}\cdot g = \dfrac{4.25\,\mathrm{kg}-2.3\,\mathrm{kg}}{4.25\,\mathrm{kg}+2.3\,\mathrm{kg}}\cdot 9.8\,\mathrm{N/kg} = 2.9\,\mathrm{m/s}^2.$