Question

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?

Answer 1

Answer:

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 = $2*10- 2x$

= $20- 2x$

Leakage = $\frac{0.1}{10} * x$

= $0.01 x$

Total  weight = $5 + 20 - 2x - 0.01 x$

= $25 - 2.01 x$

Force = wg

Force = ($25 - 2.01 x$)g

Force = 9.8 ($25 - 2.01 x$)

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

$W = \int\limits^{10}_0 {(25-2.01 x)} \, 9.8 dx \\\\W = 9.8 (25x - \frac{2.01x^2}{2}})^{10}__0$

$W = 9.8 (25*10-2.01*50)\\\\W = 1465.1 \ \ J$