Question

A 10-foot-long chain hangs from the ceiling. It weighs 45 lb. Find the total amount of work done (in ft-lb) when you raise the lower end of the chain up to the ceiling so that it is level with the upper end. Round your answer to two decimal places.

Answer 1

Workdone = 202.5ft/lb

Step-by-step explanation:

The linear density of chain = 45/10 = 4.5ft/lb

Let the ceiling be at y = 10

Then we can say that the part of chain for y <= 5does notmove only the part of the chain with y <5 gets lifted up.

Let y = distance moved by a segment of the chain

F(y) = { 10 - 2y if y > or= 5 if y < 5

Workdone = Integral (limits 0,10) Linear density × F(y) dy

Workdone = integral 4.5 × F(y) dy

W = 4.5 integral(limit 0 , 5) 10 - 2ydy + 4.5 integral(limit 0, 10) 0 × dy

W = 4.5[10y - y^2] + 0

W = 4.5[10 × 5 - 5^2] - 2.5[10× 0 - 0^2]

W = 4.5 × [50 - 25] - 0

Workdone = 4.5 × 45

Workdone = 202.5 ft/lb

Answer 2

112.50 ft-lb

Step-by-step explanation:

The linear density of the rope λ = 45 lb/10 ft = 4.5 lb/ft

Since the length of the rope is 10 ft and the bottom is raised to the ceiling, its length now becomes 10/2 = 5 ft. So, Let y = 0 define the position of the bottom of the chain and y = 10 define the position of the top of the chain. When the bottom is raised to the ceiling, its length 2y. Its position is thus 10 - 2y. So we define a function f(y) such that

`f(y)= \left \{ {{0 if y \geq } 5\atop {10 -2y if y < 5}} \right.`

Workdone = ∫₀¹⁰λf(y)dy = ∫₀⁵λf(y)dy + ∫₅¹⁰λf(y)dy

= ∫₀⁵λf(y)dy + ∫₅¹⁰λf(y)dy

= λ∫₀⁵f(y)dy + λ∫₅¹⁰f(y)dy

= 4.5∫₀⁵(10 -2y)dy + 4.5∫₅¹⁰0dy

= 4.5[10y - y²]₀⁵

= 4.5[10(5) - 5² - (10(0) - 0²)]

= 4.5(50 - 25 - 0)

= 4.5(25)

= 112.50 ft-lb