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Two masses of 60 N & 30N are positioned over frictionless & massless pulley (fig.) If the 60 N. mass is released from rest,

find the speed at which the 60N mass will hit the ground.

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For block A

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$∑Fy = m\space a_y (\text {D Alembert principle}) \\ T – 30 = (30/9.8) a \\ T – 3.06 a = 30 ….i $

For block B

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$∑Fy = m ay (\text { D Alembert principle} ) \\ T – 60 = -(60/9.8) a \\ T + 6.12 a = 60 ….ii$

Solving I and ii

$T = 40N \\ . a = 3.27 m/s^2 $

Using third kinematical equation

$V^2 = u^2 + 2as \\ V^2 = 0^2 + 2\times 3.27 \times 2 \\ V = 3.62 m/s$

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