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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

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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

Grade:12

2 Answers

Sanchit Gupta
8 Points
14 years ago

NOTE the mass of the block is 45 kg.

Badiuddin askIITians.ismu Expert
148 Points
14 years ago

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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Badiuddin

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