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Olin [163]
3 years ago
7

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?
Physics
1 answer:
iren [92.7K]3 years ago
7 0

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

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