Object falls, springs catch

  • Thread starter JoshuaR
  • Start date
  • #1
26
0
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...
 

Answers and Replies

  • #2
960
0
Theres a typo in your spring constant so can't confirm answer but work looks fine.
 
  • #3
26
0
sorry, k1=3kN/m and k2=10kN/m if I remember correctly. Thanks for the confirmation.
 

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