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) 300J​

Answer 1

Answer:

3) 200

Explanation:

Answer 2

Given: Mass of the body= 6kg, relative density= 3, depth of lake= 5m

To find: work done in lifting the body

Solution: we are given the mass of the body in air= 6kg

Relative density of the body= 3

We know that

Net force= Weight- buoyancy

Buoyancy= dw× Vg

W= mgmg= ds×Vg

Net force (f) = Vg(ds-dw)

Net force = ma= ds×Vg

a= g(1-(ds/dw))

here ds is density of body, dw is density of water and V is volume of the body.

Also ds/dw = 3

therefore a= 2g/3

Net work done against the forces will be

W= maX where X is displacement that is 5m

W= 6× 2×10×5/3

W= 200J

Therefore, work done in lifting a body of mass 6kg and relative density 3 lying in the bed of a lake of depth 5m will be 200J