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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? |
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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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