Question

A spring has a natural length of 50 cm and a force constant of $2.0\times {10}^{3}N{m}^{-1}$ . A body of mass 10 kg is suspended from it and the spring is stretched. If the body is pulled down further stretching the spring to a length of 58 cm and released, it executes simple harmonic motion. What is the net force on the body when it is at its lowermost position of its oscillation? Take $g=10m{s}^{-1}$

• A. 20 N
• B. 40 N
• C. 60 N
• D. 80 N

Answer 1

The correct option is C 60 N

The extension produced in the string when the mass is suspended from it is y = $\frac{mg}{k}=\frac{10\times 10}{2\times {10}^3}=0.05m=5cm$
Thus the equilibrium position of the mass corresponds to a position when the length of the spring = 50+5=55 cm. The lowermost position of the mass corresponds to a position is (8-5)=3 cm below the equilibrium position;. At the equilibrium position, no net force acts on the
body because the downward force mg is balanced by the restoring ky of the spring. Therefore, at the lowermost position, the net force on the mass is
F = force constant × extension below the equilibrium position
= $2\times {10}^3\times 0.03=60N(\because 3cm=0.03m)$
Hence the correct choice is (c).