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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```
8 years ago

Sanchit Gupta
8 Points
```										NOTE the mass of the block is 45 kg.
```
8 years ago
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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```
8 years ago
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