Correct Answer - Option 2 : 5000 N
Concept:
The kinetic energy of the moving mass is converted into the potential energy of the spring.
Kinetic energy = \(\frac{1}{2}m{v^2}\;\)
The potential energy of spring = \(\frac{1}{2}k{x^2}\)
Stiffness: The force required to produce unit deflection in the spring is called as the stiffness of the spring.
Calculation:
Deflection (x) = 150 mm = 0.15 m
From energy conservation,
\( \frac{1}{2}k{x^2} = \frac{1}{2}m{v^2}\\\)
\( \Rightarrow k = m{\left( {\frac{v}{x}} \right)^2} = 1500 × {\left( {\frac{1}{{0.15}}} \right)^2} = 66.667\ kN/m \)
Stiffness of one spring (k) \(= \frac{{66.667}}{2} = 33.33\ kN/m\)
Maximum force = k × x = 33.33 × 0.15 = 5 KN = 5000 N
