A leaky 10-kg bucket is lifted from the ground to a height of 13 m at a constant

Question

A leaky 10-kg bucket is lifted from the ground to a height of 13 m at a constant speed with a rope that weighs 0.6 kg/m. Initially the bucket contains 39 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 13-m level. Find the work done. (Use 9.8 m/s^2 for g.) Show how to approximate the required work by a Riemann sum. (Let x be the height in meters above the ground. Enter x_i astir as x_i.) lim n rightarrow infinity sigma^n i = 1 ()Delta x Express the work as an integral. Integral_0 () dx Evaluate the integral. (Round your answer to the nearest integer.) J

Explanation / Answer

1.   (a) Dividing up the 13 m into n sections, we have work

W = sum(k=0 to n){[F(k*(13/n))]*(13/n)},

where F(y) is the force at height y, with y = 0 at ground level.

(Recall that work is the sum of (force times distance) over all the increments of motion. Also, since the bucket is being raised at a constant rate there is no force other than the force of gravity.)

Now F(y) = m(y)*g, where m(y) is the mass of the rope plus the mass of the bucket plus the mass of the water at height y.

Thus m(y) =[(13y)(0.6)+10+(39(13-y)/13 )] = (56.8-3.6y)
  
The reason for the (13 - y)/13 factor is that this is the fraction of the water that is left in the bucket at height y.

So W = sum(k=0 to n){[F(k*(13/n))]*(13/n)} =

sum(k=0 to n){[(56.8 - (3.6)*(k*(13/n))*g]*(13/n)}.

We then let n -> infinity.

2... Suppose the bucket has already moved a distance y metres vertically. Let dW be the work done moving it an additional dy m vertically.

Then dW = (weight of rope + bucket + water)×dy = [(13y)(0.6)+10+(39(13-y)/13 )]g×dy = (56.8-3.6y)g dy.

Therefore the total work done is: W = (integral(y=0 to 13)) (56.8-3.6y)g dy = 434.2 × gJ = 4255.16 J.=4255 J

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