A 5 kg bucket is lifted from the ground into the air by pulling in 10 meters of rope with linear density of 2 kg/m at a constant speed. The bucket starts with 100 cm^3 of water and leaks at a constant rate. It finishes draining just as it reaches the 10 meter mark. How much work was spent lifting the bucket, rope, and water?

Workdone W = 1465.1 J

Explanation:

The weight of the water = density × volume

weight of the water = 1000 kg/m³ × 100 cm³

weight of the water = 1000 kg/m³ × 0.0001 m³

weight of the water = 0.1 kg

weight of the bucket  = 5 kg

weight of the rope =

=

Leakage =

=

Total  weight =

=

Force = wg

Force = ()g

Force = 9.8 ()

Finally; the amount of work spent in lifting the bucket, rope, and water is calculated as follows:

Normally, the water pressure inside a pump is higher than the vapor pressure: in this case, at the interface between the liquid and the vapor, molecules from the liquid escapes into vapour form. Instead, when the pressure of the water becomes lower than the vapour pressure, molecules of vapour can go inside the water forming bubbles: this phenomenon is called cavitation. 

So, cavitation occurs when the pressure of the water becomes lower than the vapour pressure. In our problem, vapour pressure at  is 1.706 kPa. Therefore, the lowest pressure that can exist in the pump without cavitation, at this temperature, is exactly this value: 1.706 kPa.