When a weight of 500 N falls on a spring of stiffness 0.5 kN/m from a height of 2 m. What is the maximum deflection caused in first fall?

Concept:

If we neglect the air resistance then,

By using energy conservation,

(K + P)I, system = ( K + P)f , system       …(1)

Where, K, P are representing kinetic energy and potential energy. The system includes spring and mass

Taking the reference line as shown in the figure,

\( ⇒ {\bf{mgh}} + 0 = \frac{1}{2}{\bf{k}}{{\bf{x}}^2} + {\bf{mg}}\left( { - {\bf{x}}} \right)~\)

∵ mass after hitting comes below the reference line by maximum compression (x) so potential energy for the mass will be taken as with (-) sign. Also, at extreme compression of the spring, the Kinetic energy of the mass will be zero.

\( ⇒ {\rm{mg}}\left( {{\rm{h}} + {\rm{x}}} \right) = \frac{1}{2}{\rm{k}}{{\rm{x}}^2}~\)

where m = mass, h = height, k = spring constant, x = maximum deflection in the spring

Calculation:

Given:

Weight (W) = 500 N; h = 2 m; k = 0.5 kN/m = 500 N/m

Weight (W) = mg = 500 N

\( ⇒ 500 \times \left( {2\ +\ x} \right) = \frac{1}{2} \times 500 \times {x^2}\)

⇒ x = 3.24 m

Hence the deflection caused in the first fall will be 3.24 m.

Mistake PointsThe exact answer to this question is 3.24m, but 3.24 m is not available in the options, this is the official of ISRO LPSC Technical Assistant and they have given the correct answer 2 m.

So, we have to neglect the change in potential energy due to the deflection of the spring.

Neglecting the potential energy change due to the deflection of the spring, we will get the expression:

\( ⇒ mgh= \frac{1}{2}{\rm{k}}{{\rm{x}}^2}~\)

\( ⇒ 500 \times \left( {2} \right) = \frac{1}{2} \times 500 \times {x^2}\)

⇒ x = 2 m

Hence the approximated deflection caused in the first fall will be 2 m.