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Sanchit Gupta Grade: 12
        

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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

7 years ago

Answers : (2)

Sanchit Gupta
8 Points
										

NOTE the mass of the block is 45 kg.

7 years ago
Badiuddin askIITians.ismu Expert
147 Points
										

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

7 years ago
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