Respuesta :
Answer:
The answer to the question is
W = 25 × (Building height) - 30× ( total time taken) + (Building height)²
which is of the form W = Y² + 25·Y - 30·t where Y = Building height and t = total time from the start to the top of the building
Step-by-step explanation:
Firstly let us take the height as = Y
consider the work done in lifting the bucket,
[tex]W_{cable}[/tex] = [tex]\int\limits^Y_0 {{2x, dx[/tex] = (2·Y²/2) = Y² ft·lb
Next to consider the work done in lifting the water alone we have
The weight at the top of the building = 5 lb
Weight at the start of the lift = 30 lb
The work done in lifting the bucket from level x to level x - dx is the product of the weight dx
Therefore the total work in lifting the bucket is
[tex]\int\limits^Y_0 {{Weight at level x, dx[/tex]
The position is time dependent thus x = Y - Vt ft
[tex]\int\limits^Y_0 {{Y-t*V} \, dx[/tex]
The water weight after t seconds is 30 - tl lb, where l = leak rate
where t1l = 25 lb total time =
Therefore dx = -Vdt
Substituting we have vt = Y and lt = 25 lb substituting t at
[tex]\int\limits^b_0 {} \, {{30-t*l} \, (-Vdt)[/tex]
= [30·t -t²·L·V]ᵇ₀ = -30·b +b²× l×v = b(-30 + blv) but bv = Y and bl = 25
Therefore we have → 25Y- 30b
or W = 25 × (Building height) - 30× ( total time taken) + (Building height)²
which is of the form W = Y² + 25·Y - 30·t
To solve a specific case, let height of the building be 5 ft and the cable weighs 1.5 lb/ft pull rope speed = 1 ft/s
bucket leak rate = 25 lb in 5 seconds = 5 lb/s
To calculate the work done in lifting the cable we have
considering a small length of cable dx located at x below the top has a weight of 1.5 dx, the required work to completely pull the total length of the cable to the top is dW = 1.5 x dx, that is
[tex]W_{cable} =[/tex] [tex]\int\limits^ 5_0 {\frac{1}{2} x} \, dx[/tex] = [x²/4]⁵₀ = 25/4 ft·lb
For the bucket without water we have
[tex]W_{bucket}[/tex] = 5 × 2 = 10 ft·lb
Next we calculate the work done is given by considering a displacement dx of the water from level x = weight of the water at x × dx that is the total work done is given by
[tex]W_{water} = \int\limits^5_0 {Weight of water at level x} \, dx[/tex] as the level of the water at a particular time is dependent on time we have 30 - 5·t lb and the distance to the top of the building = 5 -t substituting, we have
[tex]\int\limits^0_5 {30 - 5t \, (-dt) =[/tex] [-30·t + 5·t²/2]⁵₀ = 87.5 ft·lb
Therefore the answer is
Total work done = [tex]W_{cable} + W_{bucket} + W_{water}[/tex] = 25/4 ft·lb + 10 ft·lb + 87.5 ft·lb = 103.75 ft·lb
