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In the system shown in figure mB = 4kg, and mA = 2 kg. The pulleys are massless and friction is absent everyrwhere. The acceleration of block A  is

a
103m/s2
b
203m/s2
c
2 m/s2
d
4 m/s2

detailed solution

Correct option is A

For movable pulleyaB = O+aA2aA = 2aB--------(i)Free body diagrams of A and B: Equation of motion of block A and BFor A : T-mAg2 = mAaA = mA(2aB)-----(ii)For B : mBg-2T = mBaB-------------(iii)From (ii) and (iii) [2(ii) + (iii)]mBg-mAg = [4mA+mB]aB⇒aB = [mB-mA]g[4mA+mB] = [4-2]×10[4×2+4]           = 2012 = 53m/s2Hence  aA = 2aB = 103m/s2

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A constant force F = m2g/2 is applied on the block of mass m1 as shown. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.


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