Question

A person draw a water 5m deep well through bucket of 2kg of capacity 8lit by rope of mass 1kg .total work done by person

Answer 1

Here,

• Depth of well, $\sf{h=5m}$

• Mass of bucket, $\sf{m_b=2\ kg}$

• Capacity of bucket, $\sf{V=8\ L}$

• Density of water, $\sf{\rho=1\ kg\,L^{-1}}$

• Mass of rope, $\sf{m_r=1\ kg}$

Let $\sf{g=10\ m\,s^{-2}.}$

Work done to pull up a bucket of mass $\sf{m_b}$ and capacity $\sf{V}$ through a distance $\sf{h}$ is,

$\longrightarrow\sf{W_1=(m_b+m_w)\,gh}$

But mass of water, $\sf{m_w=\rho V=V.}$ Then,

$\longrightarrow\sf{W_1=(m_b+V)\,gh}$

$\longrightarrow\sf{W_1=(2+8)\times10\times5}$

$\longrightarrow\sf{W_1=500\ J}$

The rope tied to the bucket has the same length as the depth of the well.

So work done to pull up the rope of mass $\sf{m_r}$ hanging vertically downwards is,

$\longrightarrow\sf{W_2=\dfrac{m_rgh}{2}}$

$\longrightarrow\sf{W_2=\dfrac{1\times10\times5}{2}}$

$\longrightarrow\sf{W_2=25\ J}$

Hence total work done is,

$\longrightarrow\sf{W=W_1+W_2}$

$\longrightarrow\underline{\underline{\sf{W=525\ J}}}$