Question

Leaky Bucket A 1-kg bucket resting on the ground contains 3 kg of water. How much work is required to raise the bucket vertically a distance of 10 m if water leaks out of the bucket at a constant rate of 1/5 kg/m? Assume the weight of the rope used to raise the bucket is negligible. (Hint: Use the definition of work, W = ∫a^bF(y) dy, where F is the variable force required to lift an object along a vertical line from y=a to y=b.)

Accepted Answer

Identify the variable force function F(y) that represents the weight of the bucket plus the remaining water at height y. Since water leaks at a constant rate of $$ \frac{1}{5}  kg $$per meter, the mass of water at height y is $$ m(y) = 3 - \frac{1}{5}y  kg. $$The total mass being lifted at height y is then $$ 1 + m(y) = 1 + 3 - \frac{1}{5}y = 4 - \frac{1}{5}y  kg. $$Express the force F(y) as the weight of the bucket plus water at height y. Since weight is mass times gravitational acceleration $$ g $$, we have $$ F(y) = (4 - \frac{1}{5}y) 	imes 9.8 $$ newtons. Set up the integral for work using the formula $$ W = \int_a^b F(y) \, dy $$, where $$ a = 0  m ($$starting height) and $$ b = 10  m ($$final height). So, $$ W = \int_0^{10} (4 - \frac{1}{5}y) 	imes 9.8 \, dy . $$Simplify the integral by factoring out constants and rewriting the integrand: $$ W = 9.8 \int_0^{10} \left(4 - \frac{1}{5}yight) dy . $$Evaluate the integral by integrating term-by-term: $$ \int_0^{10} 4 \, dy $$ and $$ \int_0^{10} \frac{1}{5}y \, dy $$, then multiply the result by 9.8 to find the total work done in lifting the bucket and leaking water.

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Key Concepts

Work as an Integral of Variable Force

Force Due to Weight and Variable Mass

Modeling Mass Loss as a Function of Height