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.)

Verified step by step guidance

1. 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.
2. 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 \( F(y) = (4 - \frac{1}{5}y) \times 9.8 \) newtons.
3. 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, \( W = \int_0^{10} (4 - \frac{1}{5}y) \times 9.8 \, dy \).
4. Simplify the integral by factoring out constants and rewriting the integrand: \( W = 9.8 \int_0^{10} \left(4 - \frac{1}{5}y\right) dy \).
5. 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.

Key Concepts

Work as an Integral of Variable Force

Work done by a variable force is calculated using the integral W = ∫ F(y) dy, where F(y) changes with position y. This approach sums infinitesimal amounts of work over the distance, accounting for forces that vary as the object moves.

Force Due to Weight and Variable Mass

The force required to lift an object vertically equals its weight, which depends on mass and gravity (F = mg). When mass changes with height, as water leaks out, the force becomes a function of position, requiring careful modeling of mass variation.

Modeling Mass Loss as a Function of Height

When water leaks at a constant rate per meter, the mass decreases linearly with height. Expressing mass as a function of vertical position allows calculation of the changing force and integration over the lifting distance.