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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.
I'm not sure how to solve it, but \(answer\) comes approx = to \(11.1~kN/m\)
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.
\(P.E. = K.E. =225 J \)
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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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.