A bumper consists of two helical springs of circular section brings to rest a railway wagon of 1500 kg and moving at 1 m/sec. While doing so, the springs are compressed by 150 mm. Then, the maximum force in N on each spring (assuming gradually increasing load) is

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