The system is released frm rest with the spring initially stretched 75 mm. The spring has a stiffness of 1050 N/m. Show that the velocity of the block after it has dropped 12 mm will be 0.371 m/s. Neglect the mass of the small pulley

NOTE the mass of the block is 45 kg.

HI sanchit gupta

Here one thing shoul be noted that when block falls 12 mm then spring must be streched by 2*12=24 mm

now energy balance

initial energy = 45*g*.012 +(1/2) *1050 *(.075)²

final energ   = (1/2) *45 *v² +(1/2) *1050*(.075+.024)²

equate these two and calculate v

v=.37 m/s

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