Laws of MotionHard
Question
The pulley arrangements shown in figure are identical, the mass of the rope being negligible. In case I, the mass m is lifted by attaching a mass 2m to the other end of the rope. In case II, the mass m is lifted by pulling the other end of the rope with a constant downward force F = 2mg, where g is acceleration due to gravity. The acceleration of mass in case I is:
Options
A.zero
B.more than that in case II
C.less than that in case II
D.equal to that in case II
Solution
T1 − mg = ma1 [Newton′s II law for m]
2 mg − T1 = 2 ma1 [Newton′s II law for 2m]
⇒ a1 = g/3
Case 2
F − mg = ma2 [Newton′s II law for m]
⇒ 2 mg − mg = ma2 ⇒ a2 = g ⇒ a2 > a1
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