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anonymous
 one year ago
Need help getting started on this question. Thanks
anonymous
 one year ago
Need help getting started on this question. Thanks

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

Koikkara
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure how to solve it, but \(answer\) comes approx = to \(11.1~kN/m\)

Koikkara
 one year ago
Best ResponseYou've already chosen the best response.0The 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

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0I 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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