Find the value of acceleration of block B (in m/s2) for the system shown in figure. Take g=10m/s2.
A
152m/s2
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B
52m/s2
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C
8m/s2
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D
252m/s2
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Solution
The correct option is A152m/s2 Total length of string (l) will remain constant. Let the length of the string between block B and pulley attached to block A be xA.
∴l=xB+3xA...(i) where l represents total string length at the level of the pulley attached to A. Differentiating Eq (i) twice w.r.t time, ⇒0=d2xBdt2+3d2xAdt2 ⇒0=−aB+3aA (aB is −ve since distance xB is decreasing) ⇒aB=3aA...(ii)
Applying Newton's 2nd law for block B, T=maB....(iii) For block A, 3mg−3T=3maA or mg−T=maA...(iv)
From Eq. (ii),(iii),(iv): mg=maA+maB ⇒mg=4maA ∴aA=104=52m/s2 &aB=152m/s2