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 \(y+\Delta y\).

Short answer

The work done is 43,200 foot-pounds.

Step-by-step solution

Understanding the Problem

First, we identify that the work done is the integral of the force applied over the distance. Since the weight decreases with the loss of sand, our force is variable. We can represent this scenario over an interval from 0 to 80 feet as the bucket is lifted.

Establish the Variable Force Equation

The weight of the bucket at any point is the weight of the bucket plus the remaining sand. Initially, the total weight is 600 pounds (100 lb for the bucket and 500 lb for the sand). We know sand leaks at 3 pounds per second. Therefore, at any time t, the weight of the sand is 500 - 3t.

Calculate the Total Remaining Weight with Respect to y

Since y represents the height, and the bucket moves upwards at 2 feet per second, time can be expressed as t = y/2. Therefore, the weight of the sand at a height y is 500 - 3(y/2). The total weight at height y is 100 + (500 - 3(y/2)) pounds.

Express the Work Required for a Small Section

The incremental work \Delta W required over a small vertical segment \Delta y is the force (weight at that height) times the small vertical distance: \(100 + \left(500 - \frac{3y}{2}\right)\Delta y\).

Set up the Integral for Total Work Done

Integrate the incremental work from y = 0 to y = 80: \[ W = \int_{0}^{80} \left( 100 + 500 - \frac{3y}{2} \right) \, dy \] Simplified, we have: \[ W = \int_{0}^{80} \left( 600 - \frac{3y}{2} \right) \, dy \]

Solve the Definite Integral

Compute the integral: \[ W = \int_{0}^{80} (600 - \frac{3y}{2}) \, dy = \left[ 600y - \frac{3y^2}{4} \right]_{0}^{80} \] Evaluating this gives: \[ W = \left(600 \times 80 - \frac{3 \times 80^2}{4}\right) - (600 \times 0 - \frac{3 \times 0^2}{4}) \]\[ W = 48000 - 4800 = 43200 \]

Conclusion

Therefore, the total work done in lifting the bucket is 43,200 foot-pounds.

Key concepts

Work and Energy

Work and energy are fundamental concepts in physics, often paired together as they are directly related. The work done by a force is defined as the product of the force and the distance over which it acts. In mathematical terms, when a constant force \( F \) acts on an object to move it through a distance \( d \), the work \( W \) done is given by: \[ W = F \times d \] However, when the force is variable, work is calculated through integration. Energy is the capacity to do work, and in this context, we're dealing with mechanical energy. As the crane lifts the bucket, it applies work against the force of gravity, effectively increasing the potential energy of the bucket. This problem asks us to calculate the total work needed to lift the bucket, accounting for the variable weight due to sand leaking out.

• Understand that work equals force times distance.
• Recognize the implications of a variable force on work calculation.

Integrals and Integration

Integrals are a core component of calculus, used to calculate areas, volumes, central points, and work, among other things. In this exercise, we use integration to find the total work done by summing up infinitesimal work elements over a continuous interval. Integration allows us to cope with a variable force – in this case, the weight of the bucket changes as sand leaks out. The integration process begins with the function representing work over a small distance \( \Delta y \): \[ \Delta W = (100 + (500 - \frac{3y}{2})) \Delta y \] Next, by setting up the integral over the range from 0 to 80 feet, we encapsulate the entire process of lifting the bucket, capturing the change in force over height: \[ W = \int_{0}^{80} (600 - \frac{3y}{2}) \ dy \] Thus, integration provides a means to sum up all incremental work over the lifting range.

• Integrals help in summing up small work contributions.
• Flexible handling of variable functions.

Variable Force

When a force is not constant, it is described as variable. A variable force changes in magnitude or direction with time or position. In this problem, the force applied by the crane fluctuates because the bucket’s weight decreases as sand leaks out. This is a classic situation illustrating the need for calculus. By expressing the force as a function of height \( y \), we derive \[ F(y) = 100 + (500 - \frac{3y}{2}) \] At any height, the current force acting is the total remaining weight. This formula captures how the bucket's weight reduces as it ascends. Calculating work when dealing with such variable forces often requires integration to sum the continuous changes across the distance.

• Variable forces require dynamic calculations.
• Use of force functions to reflect changes over time or space.

Differential Equations

Differential equations play a key role in understanding dynamics involving variable forces. They allow us to relate functions and their rates of change. While this specific problem doesn't directly ask for solving a differential equation, the approach used hints at differential calculus principles. By figuring out the weight as a function of height using the sand leaking rate, we essentially use a derivative-type reasoning: \[ \text{Sand weight with respect to time } t = 500 - 3t \] By converting to height with \( t = y/2 \), we model the decrement based on both time and height. This illustrates how calculus aids in modeling dynamic systems where forces are not static.

• Differential equations describe dynamic changes in physics effectively.
• While not solved here, they underpin much of the rationale for variable force models.