An anchor weighing 105 lb in water is attached to a chain weighing 3 lb/ft in water. Find the work done to haul the anchor and chain to the surface of the water from a depth of 35 ft.

The weight of the anchor is 105 lb

The chain weights  3 lb/ft

The length of the chain to be hauled up  be y

Therefore the total work done is 5336 ft-lbs

Explanation:

The weight of the anchor is 105 lb

The chain weights  3 lb/ft

The length of the chain to be hauled up  be y

So the combined weight is

==> (3)(35-y)+100=(205-3y)

Consider a small section of chain of length Δy

The work done by a small section is (205-3y)(Δy)

Total work done to haul up the anchor and the chain can be calculated as

W=∫₀³⁵(205-3y)Δy

  =(175y-3y²/2)²⁵

  =(205(35)-3(35²-0)/2)

  =5336

Therefore the total work done is 5336 ft-lbs

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.