anonymous
  • anonymous
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chestercat
  • chestercat
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anonymous
  • anonymous
The steel ingot shown in Fig. 2 has a mass of 1800 kg. It travels along the conveyor at a speed v = 0.5 m/s when it collides with the nested spring assembly. Determine the maximum deflections in each spring needed to stop the motion of the ingot, if kA = 5 kN/m and kB = 3 kN/m.
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Koikkara
  • Koikkara
I'm not sure how to solve it, but \(answer\) comes approx = to \(11.1~kN/m\)
Koikkara
  • Koikkara
The way i found is .... The kinetic energy of the ingot is \(K.E. = 1/2 m*v^2\) = \(1/2* 1800 * 0.5^2\) = 225 J For the ingot to stop, the springs must decrease the kinetic energy until it is zero, this energy is stored as potential energy in the spring. therefore \(P.E. = K.E. =225 J \) Now to find, Potential energy stored in the springs: \(P.E. = 1/2( k1* x1^2 + k2*x2^2) \) And when the ingot is 0.3m from the wall, the compression of 2 springs : \(x1 = 0.5 - 0.3 = 0.2 m \) \(x2 = 0.45 - 0.3 = 0.15 m\), where spring 1 in the outer spring \(Substituting\), values and the spring constant for the outer spring: \(P.E. = 225 = 1/2( 5000* 0.2^2 + k2*0.15^2) \) \(= 225 = 1/2( 200 + k2*0.0225) \) rRearranging in terms of k2, \(k2 = (2*225 - 200)/0.0225 \) \(= 11.1 kN/m \) @ohernand

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mathmate
  • mathmate
I am not sure if you are solving the same problem as in the figure. In your question, ka and kb are given, and there is a 5 cm distance between the beginning of the two spring. We're expected to find the \(distance\) when the block stops. Your solution calculates k2, which seems to be a different problem with given displacements. So please clarify.

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