Question

Calculate the work done in raising a stone of mass $5kg$ and specific gravity $3$,lying at bed of a lake through a height of $5m$.

Answer 1

First we will calculate the loss of weight in water and then weight of stone in water and then we will calculate the force due to gravity and then we know to calculate work done we will produce it to the distance given in question.

Complete step by step answer:
In this question we are given with mass of stone as, $5kg$
And we are also given with specific gravity in question as, $3$
And also they have given distance that is height, $5m$
So now to calculate the loss of weight in water as, ${L_{water}} = \dfrac{{{W_{air}}}}{{S.G}}$,where $S.G$is specific gravity and ${W_{air}}$is weight of stone in air, now substitute the values in equation and we get
${L_{water}} = \dfrac{5}{3}$
And now to calculate the weight of stone in water as, ${W_{siw}} = 5 - \dfrac{5}{3}$
${W_{siw}} = \dfrac{{10}}{3}$
And now to calculate the force of gravity as, $F = \dfrac{{10}}{3} \times 10$, where $10$is acceleration due to gravity
$F = \dfrac{{100}}{3}$
Now to calculate work done we will use $W.D = F \times 4$,where distance is given in question and force is calculated above
$W.D = \dfrac{{100}}{3} \times 4$
$W.D = \dfrac{{400}}{3}$ joules.
Therefore work done is calculated.

Additional Information: Archimedes' guideline permits the lightness of any drifting article mostly or completely submerged in a liquid to be determined. The descending power on the article is just its weight. The upward, or light, power on the item is that expressed by Archimedes' standard, above. In this way, the net power on the article is the contrast between the sizes of the light power and its weight. In the event that this net power is positive, the item rises; if negative, the article sinks; and if zero, the item is impartially light—that is, it stays set up without either rising or sinking. In straightforward words, Archimedes' rule expresses that, when a body is halfway or totally drenched in a liquid, it encounters a clear misfortune in weight that is equivalent to the heaviness of the liquid uprooted by the inundated aspect of the body

Note: In this question you can make mistakes by placing the same mass to calculate the force and ignore the weight loss in water and you will get the wrong answer, so if a particle is in water first of all calculate the weight loss in water according to specific gravity given. Weight loss(apparent weight) is due to buoyant force.