- #1
JoshuaR
- 26
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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...
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...