Question: A leaky 17 kg bucket is lifted from the ground to a height of 62 m at a constant speed with a rope that weighs 0.2 kg/m. Initially the bucket contains 32 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 62-m level. How much work is done?
Loss of weight of water from bucket is 32 kg over a distance of 62 m = 32/62 kb/m = 16/31 kg/m.
Weight of bucket, including water, at distance x from ground is Wb = 17 + 32 – (16/31).x kg
Weight of rope diminishes by 0.2 kg for every metre the bucket is drawn upwards.
Initial length of rope is 62 m.
Initial weight of rope, being pulled up is, 62*0.2 = 12.4 kg
Length of rope when at a distance x m from the ground is 62 – x m.
Weight of rope when at a distance x m from the ground is Wr = (62 – x)*0.2 = 12.4 – 0.2.x kg.
Total weight being lifted at distance of x m from ground is W = Wb + Wr
W = (49 – (16/31).x) + (12.4 – 0.2.x) = 61.4 – (111/155).x
Work done over a small distance δx
δW=F∙δx
Where the lifting force, F, is just equal to the current weight, Wg.
Total work done
WD=∫dW=∫_0^x F dx=∫_0^62 Wg dx
WD=g∫_0^62 61.4 – (111/155).x dx
WD=g(61.4x - (111/155).x^2/2) [62 ..0]
WD=9.81(3,806.8 - 1376 2/5)
WD=23,842 J
WD=24KJ
A quicker way to do this problem is to notice that the rates (of loss of weight) are linear relationships. E.g. 0.2.x kg/m rather than 0.2.x^2 kg/m.
Since the relationships are linear, then just take average values and work from that.
Weight of rope = 62*0.2 = 12.4 kg.
Average weight of rope = 6.2 kg.
Average weight of water = 32/2 = 16 kg.
Weight of bucket (constant) = 17 kg.
Average weight being lifted = 17 + 16 + 6.2 = 39.2 kg
Distance over which the average weight is being lifted = 62 m.
Work done = Fx = Wg.x = 39.2*g*62 = 23,842.224 J = 24 KJ
