Answer:
Mass of hose uncoiled \[=20\times 0.2=4\,\,kg\] \[K.E.\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times 4\times {{2.2}^{2}}=9.68J\] \[\therefore \]Work done\[=9.68\,J.\]
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