Problem 17 - Calculus: Early Transcendentals

Question

Show how to approximate the required work by a Riemann sum. Then express the work as an integral and evaluate it. 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 0.8 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. How much work is done?

Step-by-Step Solution

Answer

The work done is 393.6 Joules.

Step 1: Define Variables

First, let's identify the variables. Initially, the bucket weighs 10 kg, the rope weighs 0.8 kg per meter, and starts with 36 kg of water. The water leaks at a constant rate over 12 meters, meaning the amount of water reduces to 0 kg as the bucket is lifted.

Step 2: Calculate Rope Weight Distribution

Since the rope weighs 0.8 kg/m, at any height \( x \), it contributes to the weight lifted. If \( x \) is the distance from the ground, then the amount of rope lifted is also \( x \) meters. Thus, the weight of the rope lifted is \( 0.8x \) kg.

Step 3: Water Weight Distribution

The water initially weighs 36 kg but leaks completely by 12 meters. This implies a linear decrease: at height \( x \), the water's weight is \( 36 - 3x \) kg (since \( 3x = \frac{36}{12}x \) describes the rate of leakage).

Step 4: Total Weight Calculation

Total weight lifted at a given height \( x \) is the sum of the bucket (constant 10 kg), the rope weight \( 0.8x \) kg, and the water weight \( 36 - 3x \) kg. This gives \( 10 + 0.8x + 36 - 3x = 46 - 2.2x \) kg.

Step 5: Riemann Sum Approximation

The work \( W \) done in lifting to each very small height \( \Delta x \) is given by the weight at that height \( \times \Delta x \). Thus, \( \sum (46 - 2.2x) \Delta x \) approximates the total work.

Step 6: Formulate Integral Expression

Work is the integral of force over distance: \( W = \int_0^{12} (46 - 2.2x) \, dx \). This represents summing up all the small work contributions from \( x = 0 \) to \( x = 12 \).

Step 7: Evaluate the Integral

\[ W = \int_0^{12} (46 - 2.2x) \, dx = [46x - 1.1x^2]_0^{12} \]Evaluate this from 0 to 12:\[ W = (46 \times 12 - 1.1 \times 12^2) - (46 \times 0 - 1.1 \times 0^2)\]\[ = 552 - 158.4 = 393.6 \, \text{Joules} \]

Key Concepts

Work Done

Understanding work done in physics involves figuring out how much energy is required to move an object over a distance. In our leaky bucket problem, this means calculating the energy needed to lift the bucket and rope from the ground to 12 meters high. The concept of work in physics is defined as the product of force and distance. Here, the force is due to the total weight of the bucket, water, and rope, all of which change as the bucket rises.
- **Force**: In this specific exercise, the force is the weight, which changes with height because the water leaks out at a constant rate, and the length of the rope being lifted increases.
- **Distance**: The distance here is 12 meters, but since the weight changes with height, it's vital to evaluate the force over each small segment of this distance.
Calculating work done requires summing up the small amounts of work (force times small distance increments) necessary to lift the bucket at each incremental height. This approach is formalized in the concept of a Riemann sum, which allows us to approximate the total work done with differential calculus tools.

Integral Calculus

Integral calculus plays a key role in calculating the total work done when forces vary over a distance. It helps sum up infinitely small changes to find a total. This is particularly useful when working with changing forces, like our leaky bucket case.
In the problem, the force required changes as more of the rope is lifted and as water leaks out. To accurately calculate the work done, the force must be integrated over the entire distance that the bucket travels. This means transforming the concept of discrete Riemann sums into a continuous integral.
The integral we evaluate is \( W = \int_0^{12} (46 - 2.2x) \, dx \) where the integrand, \( 46 - 2.2x \), represents the changing weight of the system as the bucket is lifted. Solving this integral from 0 to 12 effectively adds up all the small amounts of work done at every tiny interval of distance.

Variable Weight Distribution

Understanding variable weight distribution is crucial when evaluating how the weight of an object varies with height or distance, which directly affects the force needed to move it. In our exercise, several components have weight changes:

• The bucket itself has a constant weight of 10 kg, so it does not contribute to variable distribution.
• The rope's weight depends linearly on how much rope has been lifted off the ground. Since the rope weighs 0.8 kg per meter, it adds 0.8 kg for each meter lifted.
• The water leaks steadily, reducing from 36 kg to 0 kg as the bucket is raised the full 12 meters. This is a uniform decrease, which means at any height \( x \), the water weighs \( 36 - 3x \) kg.

When calculating work done over a changing weight system, like this leaking bucket scenario, it is essential to precisely account for how weight gets redistributed over the lifting distance. This variable distribution impacts the total required force as it changes continuously as the bucket ascends.