given: \(m=1.5 kg\),
\(f=32 N , e=0.35\)
\(\max K . E = m\)
we can also show that
\(K \cdot E_{\max }=\frac{1}{2} K A^{2}\).
Where \(k\) is a constant of proportionality known as elastic constant
from \(\omega=\sqrt{\frac{k}{m}}\) and \(1-\omega . A\)
thus, \(v=\sqrt{\frac{h}{m}} A\)
and \(K \cdot E=\frac{1}{2} m \cdot \sqrt{\frac{h}{m} \cdot 4}\)
\(=\frac{1}{2} \cdot m_{1} \cdot \frac{k}{m_{1}}+=\frac{1}{2} h .4\)lso \(F=h\)
\begin{aligned}
&k=\frac{f}{e}=\frac{32}{0.35}=91.428 N / m\\
&\text { for } \max K . E, A=0.2 m\\
&\therefore K . E=1 / 2 \times 91.42857 \times 0.2^{2}\\
&=1.83 J\\
&\text { Alternatively, }\\
&K \cdot E_{\max }=\frac{1}{2} m v_{\max }^{2}\\
&=\frac{1}{2} k A^{2}\\
&\text { where } K=\frac{F}{e}=\frac{32}{0.35}\\
&\text { for maximum } K . E . \text {, }\\
&A=0.2 m\\
&\text { K.E. }=\frac{1}{2} \times \frac{32}{0.35} \times 0.22\\
&=1.83 J
\end{aligned}