Respuesta :
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
The work done to haul the anchor and chain to the surface of the water from a depth of 35 ft is 5512.5 lb-ft.
What is work done?
Work done is the amount of force needed to displace an object by a distance ds. It is given by the formula,
[tex]W =\int F ds[/tex]
We know that the weight of the anchor in the water is 105 lb, while the wight of the chain is 3 lb\ft, so the total weight that is needed to be held can be given as,
Weight of the anchor + Weight of the chain that is underwater
105 lb + (3lb/ft for length (35-s))
= 105 + 105 - 3s
= 210 - 3s
Therefore, the force that will be needed to pull the chain up is (210-3s).
We know that we need to haul the chain from 35 ft inside the water to 0 ft(on the surface) of the water.
The work done to haul the chain and anchor,
[tex]W =\int_0^{35} (210-3s) ds\\\\W = [210s - \dfrac{3s^2}{2}]_0^{35}\\\\W = 5512.5\rm \ lb-ft[/tex]
Hence, the work done to haul the anchor and chain to the surface of the water from a depth of 35 ft is 5512.5 lb-ft.
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