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PhysicsW.E.P.Calculation of workMedium2 minQB

PhysicsMediumsingle choice

The system shown is released from rest. Speed of block A, after block B has descended by 2 cm, is :-(take $g = 10 m / s^{2}$ )

Answer:

Solution:

Energy balance while B descends 2 cm:

$m_{B} g h - μ m_{A} g h = \frac{1}{2} (m_{A} + m_{B}) v^{2}$

$(3 \times 10 \times 0.02) - (0.5 \times 2 \times 10 \times 0.02) = \frac{1}{2} (2 + 3) v^{2}$

$0.6 - 0.2 = 2.5 v^{2} \Rightarrow v^{2} = 0.16 \Rightarrow v = 0.4 m/s$

Ans. (A)

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Calculation of work

⏱ 2mℹ️ Source: QB

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