Question

Force constant of a weightless spring is $16\text{N/m}$. A body of mass $1.0\text{kg}$ suspended from it is pulled down through $5\text{cm}$ from its mean position and then released. The maximum kinetic energy of the body will be

• A. $2\times {10}^{-2}\text{J}$
• B. $4\times {10}^{-2}\text{J}$
• C. $8\times {10}^{-2}\text{J}$
• D. $16\times {10}^{-2}\text{J}$

Answer 1

The correct option is A $2\times {10}^{-2}\text{J}$

Let us assume , spring- mass system is initially in equilibrium at $x=0$. Let this point be the reference point about which the body executes SHM.
Given,
Mass of the body $m=1.0\text{kg}$
Elongation of spring/Amplitude of SHM $A=5\text{cm}$
Maximum kinetic energy of the body attached to the spring is given by $K_{max}=\frac{1}{2}k{A}^2$
$=\frac{1}{2}\times 16\times (5\times {10}^{-2}{)}^2=2\times {10}^{-2}\text{J}$
Thus, option (a) is the correct answer.