    Question

# Two men of masses m1 and m2 hold on the opposite ends of a rope passing over a frictionless pulley and are initially separated by 5 m . Both the men are at rest, the man of mass m1 climbs up the rope with an acceleration of 1.2 m/s2 relative to the rope. The man of mass m2 climbs up the rope with an acceleration of 2 m/s2 relative to the rope. If m1=40 kg and m2=60 kg, A
tension in the rope is 556.8 N
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B
tension in the rope is 586.8 N
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C
acceleration of rope is 2.72 m/s2
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D
acceleration of rope is 3.72 m/s2
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Solution

## The correct option is C acceleration of rope is 2.72 m/s2 Let am1G and am2G be the accelerations of man m1 and man m2 with respect to the ground. And am1R and am2R be the accelerations of man m1 and man m2 with respect to the rope. aRG is the acceleration of the rope with respect to the ground. ​​​​​​​ For man of mass m1 am1G=am1R+aRG ⇒am1G=(1.2+a) For man of mass m2, am2G=am2R+aRG =(2−a) So now, T−mg=m1(1.2+a).......(i) T−mg=m2(2−a)........(ii) Solve eq. (i) & (ii) and put m1=40 kg and m2=60 kg, you get a=2.72 m/s2 T=556.8 N  Suggest Corrections  0      Explore more