Calculating Velocity of a Block Shot from a Spring Gun

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SUMMARY

The discussion centers on calculating the velocity of a block shot from a spring gun, with specific parameters including a block mass of 0.75 kg, a spring constant (k) of 28 N/m, and a table height of 1.04 m. The correct approach involves using energy conservation principles, specifically the equation (1/2)kx^2 = (1/2)mv^2, while accounting for gravitational potential energy. The accurate final velocity when the block hits the ground is determined to be 4.73 m/s, correcting the initial miscalculation of 4.3 m/s.

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Aggie
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A block, mass 0.75 kg, is "shot" from a spring gun from a table. The table has a height 1.04 m and is frictionless. If the spring has k=28 N/m and is originally compressed 0.23 m, how fast will the block be goign when it hits the ground?


(1/2)*kx^2 = (1/2)mv^2 ?



I got 4.3 m/s but the answer is 4.73 m/s
 
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If you show us your work, we can say where you've gone wrong.
 
(23*0.23^2)/2= (0.75*9.8*1.04)+(0.7*v^2)/2
 
Hint: you have to "count in" the height of the table somehow in your calculation.
 
Aggie said:
(23*0.23^2)/2= (0.75*9.8*1.04)+(0.7*v^2)/2
Are you sure?? You need to find the velocity at the bottom and not at 1.04m. Also, k = 28N/m, not 23.
 

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