Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

A bucket weighing 100 pounds is filled with sand weighing 500 pounds. A crane lifts the bucket from the ground to a point 80 feet in the air at a rate of 2 feet per second, but sand simultaneously leaks out through a hole at 3 pounds per second. Neglecting friction and the weight of the cable, determine how much work is done. Hint: Begin by estimating \(\Delta W\), the work required to lift the bucket from \(y\) to \(y+\Delta y\).

Step-by-Step Solution

Answer

The work done is 43,200 foot-pounds.

Step 1: Understand the Work Formula

Work done by a force is calculated using the formula \( W = F imes d \), where \( W \) is work, \( F \) is force, and \( d \) is distance. In this problem, because the weight of the sand decreases as it leaks, we must express force as a function of height.

Step 2: Determine the Expression for Weight

The bucket itself weighs 100 pounds, and initially, the sand weighs 500 pounds, totaling to a 600-pound force. As the bucket is lifted, sand leaks out at 3 pounds per second. Given the lifting rate is 2 feet per second, the sand leaks \(1.5\) pounds per foot (\(\frac{3 \text{ pounds/second}}{2 \text{ feet/second}} = 1.5\text{ pounds/foot}\)). Thus, the weight of the system at height \(y\) is \( 600 - 1.5y \).

Step 3: Setup the Integration for Work Done

The work done to lift the bucket from height \(y=0\) to \(y=80\) is the integral of the force, which varies as the sand leaks out, over the distance. So the expression is:\[ W = \int_0^{80} (600 - 1.5y) \, dy \]

Step 4: Solve the Integral

Calculate the integral to find the total work. \[ \begin{align*} W & = \int_0^{80} (600 - 1.5y) \, dy \& = \left[ 600y - 0.75y^2 \right]_0^{80} \& = (600 \times 80 - 0.75 \times 80^2) - (600 \times 0 - 0.75 \times 0^2) \& = 48000 - 4800 \& = 43200 \text{ foot-pounds}\end{align*} \]

Step 5: Interpret the Result

The integral's result, 43,200 foot-pounds, represents the total work done by the crane to lift the leaking sand bucket to a height of 80 feet while accounting for the decrease in weight due to sand leakage.

Key Concepts

Integration in Calculus

Integration is a fundamental concept in calculus used to find accumulations of quantities, such as areas under a curve or total quantities over time. In our exercise, integration helps in calculating the total work done when moving a bucket with sand over a certain distance.
When a force varies with position, as in this problem where the weight decreases due to leaking sand, integration allows us to keep track of these changes. We express the total work done, \( W \), as the integral of the varying force over the height from 0 to 80 feet:

• Identify the starting setup: weight changes as sand leaks.
• Formulate the integral based on this change: \( W = \int_0^{80} (600 - 1.5y) \, dy \).

Integration captures the sum of all infinitely small bits of work done throughout the ascent.
It's a powerful tool because it translates variable problems into a format where algebraic calculus can come into play.

Understanding Variable Force

A variable force is one that changes magnitude as an object moves. In physics calculations, accurately accounting for force variations is vital as these can impact the overall result. In our example, while the crane lifts the bucket, the sand leaks out causing a steady decrement in weight, and therefore force, over time and height. Calculating the force at any given moment involves understanding how quickly the sand leaks out in relation to how fast the bucket is lifted.

How to Determine Variable Force

Here's how:

• The bucket starts with a combined force of 600 pounds (bucket + sand).
• The sand leaks at a rate of 3 pounds per second.
• Given the lift is 2 feet per second, 1.5 pounds of sand leaks out with every foot of ascent (\( \frac{3}{2} \)).

Thus, at any height \( y \), the force is \( 600 - 1.5y \) pounds.
Recognizing and calculating with a variable force is crucial in many physics-based applications, making integration a necessary skill for analyzing such dynamic systems.

Physics Applications of Work-Energy Principle

The Work-Energy Principle is a crucial concept in physics, establishing a relationship between the work done on an object and its energy changes. This principle states that work done by external forces changes an object's energy, allowing us to analyze how mechanical energy is transformed. In this exercise, work is done against the gravitational force to elevate the bucket, which causes a change in gravitational potential energy. The work-energy principle ensures that the work done, calculated as 43,200 foot-pounds, matches the net energy change in the system, considering all dynamic variations like sand leakage.
Having reliable physics applications means understanding the nuanced ways energy is added and removed from a system:

• Work increases potential energy as the bucket rises.
• The leaking sand alters how much force remains as energy is converted.

Consequently, comprehending the Work-Energy Principle lets us better grasp how systems evolve with varying forces as seen in our example.

Work-Energy Principle and Calculus

In calculus, integrating the concept of the Work-Energy Principle enhances our understanding of energy transformations. The principle leverages calculus to quantify how variable forces affect mechanical work and energy balance. In our scenario, using calculus through integration allowed us to create a precise computation of work done as the bucket ascends. Calculus tools interpret physical principles by:

• Exact calculation of variable forces acting over infinitesimal distances.
• Application of a mathematical model of changing weights due to sand leakage.
• Deriving the right expectations about energy needs for an accurate depiction.

Thus, this principle solidifies how physics applies not only in theory but in executable mathematical steps, enhancing comprehension and predicting real-world behavior.