How High Was the Plunger Released to Compress the Springs?

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SUMMARY

The discussion focuses on calculating the height from which an 8-kg plunger is released to compress two nested springs. The outer spring has a constant of k1 = 3 kN/m, and the inner spring has a constant of k2 = 10 kN/m. The maximum deflection of the outer spring is 150 mm, leading to the equation mg(h + 0.15 m) = 0.5k1(0.15 m^2) + 0.5k2(0.06 m^2). The calculated height h is determined to be 0.509 m, confirming the approach is correct despite an initial typo in the spring constant.

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JoshuaR
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1. An 8-kg plunger is released from rest in the position shown and is stopped by the two nested springs; the constant of the outer spring is k1 = 3kN/m and the constant of the inner spring is k2 = 1-kN/m. If the maximum deflection of the outer spring is observed to be 150mm, determine the height h from which the plunger was released.
Given a diagram. h is the height above the tallest spring, the outer spring. The outer spring is 90 mm higher than the inner spring.



2. mgh=PE .5kx^2=PE spring



3. mg(h+.15m) = .5k1(.15m^2) + .5k2(.06m^2)
Yielding an h of 0.509m.
Can someone inform me if this is the right path? It seems too simple...
 
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there's a typo in your spring constant so can't confirm answer but work looks fine.
 
sorry, k1=3kN/m and k2=10kN/m if I remember correctly. Thanks for the confirmation.
 

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