Question

A weighing machine consists of a 2 kg pan resting on a spring having linear characterises. In this condition of resting on the spring, the length of spring is 200 mm. When a 20 kg mass is placed on the pan, the length of the spring becomes 100 mm. The un-deformed length L in mm and the spring stiffness k in N/m are
1. L = 220, k = 1862
2. L = 200, k = 1960
3. L = 210, k = 1960
4. L = 200, k = 2

Answer 1

Correct Answer - Option 3 : L = 210, k = 1960

Concept:

The spring stiffness is given as,

\(k = \frac{{{F_2} - {F_1}}}{{{L_1} - {L_2}}}\)

where, F₁ = weight of pan, F₂ = total weight of pan and extra mass, L₁ = initial length of spring, L₂ = final length of spring

The un-deformed length of spring is, Deflection of spring due to 2 kg pan + original length of spring

Calculation:

Given:

\(k= \frac{{\left( {22 - 2} \right) \times 9.81}}{{\left( {200 - 100} \right) \times {{10}^{ - 3}}}} = 1962~N/m\)

Deflection of spring due to 2 kg pan

\(\delta= \frac{{2 \times 9.81}}{{1962}} = 0.01~m = 10\;mm\)

∴ Unperformed length = 200 + 10

= 210 mm