1.

The pulley arrangement of figure are identical. The mass of the rope is negligible. In (a) mass m is lifted up by attaching a mass (2m) to the other end of the rope. In (b), m is lifted up by pulling the other end of the rope with a constant downward force F = 2 mg. In which case, the acceleration of m is more?

Answer»

Case (a) :

a = \(\frac{2m - m} {2m + m} g\) = a = \(\frac{g}{3}\)

Case (b):

FBD of mass m 

ma’ = T – mg 

ma’ = 2 mg – mg 

⇒ ma’ = mg 

a’ = g 

So in case (b) acceleration of m is more.



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