Two masses M1=5 kg and M2=10 kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be

Question

Two masses M1=5 kg and M2=10 kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be (A) g. (B) g / 2 (C) g / 3 (D) g / 4

Tardigrade Admin · Accepted Answer

Since M2>M1, therefore M2 moves downwards and M1 moves upwards with an acceleration a as shown in the figure. ummellyme

The equation of motion for M1 is, T-M1 g=M1 a ldots( i ) Free body diagram of M2, The equation of motion for M2 is M2 g-T=M2 a Adding (i) and (ii) , we get a=((M2-M1) g/M2+M1)=((10-5) g/(10+5))=(g/3)

Q. Two masses $M_{1} = 5 k g$ and $M_{2} = 10 k g$ are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be

$g$ .

$g /2$

$g /3$

$g /4$

Q. Two masses M1​=5kg and M2​=10kg are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be

g.

g/2

g/3

g/4

Q. Two masses $M_{1} = 5 k g$ and $M_{2} = 10 k g$ are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be

$g$ .

$g /2$

$g /3$

$g /4$