Question

The force constant of weightless spring is $16\, N / m$. A body of mass $1.0 \,kg$ suspended from it is pulled down through $5\, cm$ and then released. The maximum kinetic energy of the system (spring $+$ body) will be

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

Answer 1

Answer: (A)

$2 \times 10^{-2}\, J$

Answer 2

Given, $k=16\, N / m, m=1\,k g$ amplitude $a=5 \,cm =5 \times 10^{-2} \,m$. Kinetic energy of a body executing SHM. $=\frac{1}{2} m \omega^{2}\left(a^{2}-y^{2}\right)$ When displacement $y=0$, then $KE$ is maximum. $\therefore$ Maximum $K E=\frac{1}{2} m \omega^{2} a^{2}$ $=\frac{1}{2} m\left(\frac{2 \pi}{T}\right)^{2} a^{2}$ $=\frac{1}{2} m\left(\sqrt{\frac{k}{m}}\right)^{2} a^{2}$ $=\frac{1}{2} m \times \frac{k}{m} \times a^{2}=\frac{1}{2} k a^{2}$ $=\frac{1}{2} \times 16 k t\left(5 \times 10^{-2}\right)^{2}$ $=200 \times 10^{-4}=2 \times 10^{-2}\, J$