$\frac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=10 \log \frac{(\sqrt{\mathrm{I}_{1}}+\sqrt{\mathrm{I}_{2}})^{2}}{(\sqrt{\mathrm{I}_{1}}-\sqrt{\mathrm{I}_{2}})^{2}}=10 \log \frac{\left(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}+1\right)^{2}}{\left(\frac{\mathrm{a}_{1}}{\mathrm{a}_{1}}-1\right)^{2}}$
$=20 \mathrm{\,dB}$
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$A$: standing on the horizontal surface
$B$: standing on the block
To an observer $A,$ the net work done on the block is
