Question

Two blocks of masses 10g and 30g are fastened to the ends of a cord which passes over a frictionless pulley the 10 g block rest on the table what minimum force must be applied on the 10 g block to keep it on the table what will be the tension in the court under this condition what will be the acceleration of the system and the tension in the code when the force is withdrawn

Answer 1

For two blocks of masses fastened to the ends of a cord which passes over a frictionless pulley we have as per Newton's Second law :

motion for 30g or 0.03 kg mass will be , 0.03 g - T = 0.03 a ---------(1)

motion for 10g or 0.01 kg mass will be , T - 0.01 g = 0.01 a ---------(2)

Now the answer for the third part of the question can be obtained by solving (1) & (2) and putting g = 9.8 m/s^2

a = 4.9 m/s^2 and T = 0.147 N

for the first and second part of the question the eqn (2) has to be modified as under to accommodate minimum force, t, say to be applied . 't' will be applied downward to keep the 0.01kg mass from rising up

Also now the system will be at rest and so a = 0

The eqns will be now as under :  

0.03 g - T = 0.03 x0 = 0 ----------------(3)

T - 0.01 g - t = 0.01x0 = 0 --------------(4)

solving (3) & (4) we get ,

t = min force to be applied = 0.196 N

T = tension = 0.294 N

Answer 2

Answer:

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