A person draws water from a 5,m deep well usi | vedclass

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(B) The total work done is the sum of the work done to lift the bucket,the water,and the rope.Mass of bucket $(m_b)$ = $2\,kg$.Mass of water $(m_w)$ =

Vedclass · Accepted Answer

$525$

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(B) The total work done is the sum of the work done to lift the bucket,the water,and the rope.Mass of bucket $(m_b)$ = $2\,kg$.Mass of water $(m_w)$ = $8\,litre 	imes 1\,kg/litre = 8\,kg$.Mass of rope $(m_r)$ = $1\,kg$.Depth $(h)$ = $5\,m$.Work done on bucket: $W_b = m_b \cdot g \cdot h = 2 	imes 10 	imes 5 = 100\,J$.Work done on water: $W_w = m_w \cdot g \cdot h = 8 	imes 10 	imes 5 = 400\,J$.Work done on rope: Since the rope is uniform,its center of mass is at $h/2 = 2.5\,m$. Thus,$W_r = m_r \cdot g \cdot (h/2) = 1 	imes 10 	imes 2.5 = 25\,J$.Total work $W = W_b + W_w + W_r = 100 + 400 + 25 = 525\,J$.

$A$ person draws water from a $5\,m$ deep well using a bucket of mass $2\,kg$ and capacity $8\,litre$ with a rope of mass $1\,kg.$ What is the total work done by the person? (Assume $g = 10\,m/s^2$ and the density of water is $1\,kg/litre$.)

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