A bucket filled with water has mass 32 kg. It is lifted at a constant rate, by a

Question

A bucket filled with water has mass 32 kg. It is lifted at a constant rate, by a mechanical winch, from the bottom of a well that is 18 m deep. Compute the work required to lift the bucket from the bottom of the well to the top under each of the following circumstances:

Find the work, assuming that the chain attaching the winch to the bucket is weightless.

Find the work, assuming that the chain has mass 830g per metre.

Find the work in the case where the bucket has developed a hole and water is leaking out at a constant rate, and the bucket and water combined have a mass of 16 kg when it reaches the top. Continue to assume that the chain has mass 830g per metre.

Explanation / Answer

Work Done is = (Force applied by(on) an object) * (Distance travelled)

a>

When we assume that the chain attached is weightless then the force acting on the water filled
   bucked is only due to the weight of the (bucket + water)

Now we are given that water has mass, m = 32 kg
And the well is , d = 18 m deep

=> The force acting on the bucket would be, F = mg = 32g Newton

Hence the work done to lift the bucket from the bottom to the top of the well is, W

and W = F*d = 32g*18 = 576g Joules , Here g is acceleration due to gravity

b>

Now when the chain has a mass of = 830grams per meter = .83 kg/m , this means that for per meter length
the weight of the chain is .83 kg

If 'x' is the distance in meters that the chain has to move to reach the top of the well

Then, Total weight of the chain is = .83(18-x)g Newton

=> The total weight of the system, dF = Weight of bucket + weight of chain
                                      = 32g + 0.83(18-x)g
                                      = (46.94 - 0.83x)g Newtons

Now, if we integrate dF , for x E [0 , 18] then we would get the total work done to lift the
bucket to the top.

=> integral dFdx = integral (x=0 to 18) [(46.94 - 0.83x)g]dx

            W    = g(46.94x - 0.415x^2)(x=0 to 18)
                 = g[(46.94(18) - 0.415(18)^2)]
                 = 710.46g Joules

Hence the required work done is = 710.46g Joules , g is the acceleration due to gravity

c>

Here due to a hole in the bucket water is leaking out of the bucket at a constant rate.

Now we know that by the time the bucket reacher the top it mass is = 16 kg

This means, while moving from the bottom to the top that is on travelling 18 m distance the bucket
goes from weighing 32 kg to 16 kg.

=> Rate of loss of water is = 16/18 = 8/9 kg/m

Now total weight becomes, dF = Weight of bucket + weight of chain - weight of water leaking
                             = 32g + 0.83(18-x)g - 8xg/9
                             = (46.94 - 0.83x)g - 0.889gx  
                             = (46.94 - 1.719x)g Newtons

Now, if we integrate dF , for x E [0 , 18] then we would get the total work done to lift the
bucket to the top.

=> integral dFdx = integral (x=0 to 18) [(46.94 - 1.719x)g]dx
                 
               W = g(46.94x - 0.8595x^2) (x=0 to 18)
                 = g(46.94(18) - 0.8595(18)^2)
                 = 566.442g joules


Hence the required work done is = 566.442g Joules , g is the acceleration due to gravity

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