Question

The cable is being used to lift a 2lb leaky bucket that is initially filled with 30 lb of water. When the bucket reaches the top of the building, it only has 5 lb of water left in it. If the cable was being pulled up at a constant rate of speed, how much work is done?

Answer 1

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,

$W_{cable}$ = $\int\limits^Y_0 {{2x, dx$ = (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

$\int\limits^Y_0 {{Weight at level x, dx$

The position is time dependent thus x = Y - Vt ft

$\int\limits^Y_0 {{Y-t*V} \, dx$

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

  $\int\limits^b_0 {} \, {{30-t*l} \, (-Vdt)$

= [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

$W_{cable} =$   $\int\limits^ 5_0 {\frac{1}{2} x} \, dx$ = [x²/4]⁵₀ = 25/4 ft·lb

For the bucket without water we have

$W_{bucket}$ = 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

$W_{water} = \int\limits^5_0 {Weight of water at level x} \, dx$  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

$\int\limits^0_5 {30 - 5t \, (-dt) =$ [-30·t + 5·t²/2]⁵₀ = 87.5 ft·lb

Therefore the answer is

Total work done =  $W_{cable} + W_{bucket} + W_{water}$ = 25/4 ft·lb + 10 ft·lb + 87.5 ft·lb = 103.75 ft·lb