Question

How much work is done in lifting a body of mass 6kg and relative density 3 lying in the bed of a lake of depth 5m. 1) 30J 2) 60J 3) 200 4) 300

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Answer 1

Given:

m=6kg

$g=10 \mathrm{~ms}^{-2}$

net force =W− boyancy      

ds , dw are density  of stone & water respectively & V  is volume of stone then

$\begin{array}{l}\text { boyancy }=\mathrm{d}_{\mathrm{w}} \mathrm{V}_{\mathrm{g}} \\\\\\mathrm{W}=\mathrm{mg}=\mathrm{d}_{\mathrm{s}} \mathrm{V}_{\mathrm{g}}\end{array}$

$\begin{array}{l}\mathrm{F}_{\mathrm{n}} \mathrm{et}=\mathrm{ma}=\mathrm{d}_{\mathrm{s}} \mathrm{V}_{\mathrm{g}} \\\\\mathrm{a}=\mathrm{g}\left(1-\mathrm{d}_{\mathrm{w}} / \mathrm{d}_{\mathrm{s}}\right)\end{array}$

$\text { so }, \mathrm{a}=\mathrm{g} / 2$

$\text { now work is done against this force }$

work = maX   (X is displacement)

$work=6\times\frac{g}{2}\times5$

work=3×10×5

work=150J

Hence the answer is 150J

 

Answer 2

Given:

m=6kg

g = 10 m/$s^{-2}$

Formula,  

net force =W− boyancy      

Where,

ds = density  of stone

dw = density  of water

V =  is volume of stone

boyancy = $d_{w} V_{g} \\W=mg\\a=g(1-dw/ds)\\so ,\\a=\frac{g}{2}$

work = maX   (X is displacement)

=  6 x $\frac{10}{2}$ x5

work  =  3×10×5

Hence, work done is 150J