KinematicsHard
Question
Two pulley arrangements of inure given are identical.The mass of the rope is highlighter. In fig (a), the mass m is lifted by attaching a mass 2m to the other end of the rope. In fig (b), m is lifted up by pulling the other end of the rope with a constant dotard force F = 2mg . The acceleration of m in the two case are respectively
Options
A.3g, g
B.g/3, g
C.g/3,2g
D.g, g/3
Solution
let a and a be the acceleration in both cases respectively. Then for fig (a),
(A)
T - mg = ma ...(i)
and 2mg - T = 2mg ...(ii)
Adding (i) and (ii) we get
mg = 3ma
∴ a =
For fig (b),
T′ - mg = ma′ ...(iii)
and 2mg - T′ = 0 ...(iv)
Solving (iii) and (iv)
a′ = g
∴ a = and a′ = g
(A)
T - mg = ma ...(i)
and 2mg - T = 2mg ...(ii)
Adding (i) and (ii) we get
mg = 3ma
∴ a =
For fig (b),
T′ - mg = ma′ ...(iii)
and 2mg - T′ = 0 ...(iv)
Solving (iii) and (iv)
a′ = g
∴ a =
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