Question

a stone of mass 10 kg having relative density 4 is lying at the bottom of Lake find the work done in lifting stone through height of 20 m inside the lake

Answer 1

maybe it is wrong,but please suggest thank u

Answer 2

The work done in lifting stone through height of 20 m inside the lake is 1500 J.

Explanation:

Given the mass of the stone, m = 10 kg.

The height h = 20 m.

The relative density,$\frac{d_s}{d_w} =4$

The net force acting on the stone,

F = weight of the stone - buoyant force by water.

$F = mg -d_wgV$

Here, V is the volume.

As $m = d_sV$

So,

$F = d_sVg -d_wgV$

Net force, $F = ma = d_sVa$

so above equation becomes,

$a=g(1-\frac{d_w}{d_s} )$

substitute the values, we get

$a=10m/s^2\times(1-\frac{1}{4})$

$a=7.5m/s^2$

Therefore net force, F =10 kg x 7.5 m/s^2 =75 N.

The work done is given as

W = F.d.

W = 75 N x 20 m

W = 1500 J.

Thus, the work done in lifting stone through height of 20 m inside the lake is 1500 J.

#Learn More: work done, buoyancy force.

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