A leaky 10-kg bucket is lifted from the ground to a height of 12 m at a constant speed with a rope that weighs 0.5 kg/m. Initially the bucket contains 36 kg of water, but the water leaks at a constant rate and finishes draining just as the bucket reaches the 12-m level. Find the work done. (Use 9.8 m/s2 for g.) Show how to approximate the required work by a Riemann sum. (Let x be the height in meters above the ground. Enter xi* as xi.)

The work done is 3645.6 J or 3.6 kJ

Step-by-step explanation:

M1 = (0.5 kg/m)(12 - x m) = (6 - 0.5x) kg

At the height of x meters (0<=x<=12), the mass of the rope is,

M2 = (36kg/12m)(12 - x m) = (36 - 3x m) kg

At the height of x meters (0<=x<=12), the mass of the water is,

M3 = 10 kg

The mass of the bucket is 10 kg

Then, Total mass M is,

M = M1 + M2 + M3 = (6 - 0.5x) + (36 - 3x) + 10 = (52 - 3.5x) kg

From Force F = mass m * acceleration due to gravity g

Force = mass x gravity

F = 9.8(52 - 3.5x) Newtons

To get the work done W,

Follow the attached file,

The work done is 3645.6 J or 3.6 kJ

u rich

step-by-step explanation:

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step-by-step explanation:

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