Question

A \(100 \mathrm{lb}\) bag of sand is lifted uniformly \(120 \mathrm{ft}\) in one minute. Sand leaks from the bag at a rate of \(1 / 4 \mathrm{lb} / \mathrm{s}\). What is the total work done in lifting the bag?

Short answer

The total work done is 11100 foot-pounds.

Step-by-step solution

Understand the Problem

The total work done in lifting the bag includes both the action of lifting the 100 lb bag and the effect of the sand leaking. Our aim is to calculate the total work required. The bag loses weight as sand leaks out, so the force changes over time.

Determine the Rate of Weight Loss

Since the sand leaks at a rate of \( \frac{1}{4} \mathrm{lb/s} \), over the span of one minute (which is 60 seconds), the total weight of the sand lost is \( 60 \times \frac{1}{4} = 15 \mathrm{lb} \). This implies that the final weight of the sand in the bag, at the final position, is \( 100 - 15 = 85 \mathrm{lb} \).

Set Up the Work Calculation

The work done is equal to the integral of force over distance. Since the force is variable due to the leakage, the instantaneous force at time \( t \) (in seconds) is \( 100 - \frac{1}{4}t \). The bag is being lifted at a uniform rate, so \( x(t) = 2t \) and \( dx = 2dt \).

Calculate Work via Integration

The work done is given by:\[W = \int_{0}^{60} F(x) \, dx = \int_{0}^{60} (100 - \frac{1}{4}t) \, 2 \, dt\]This evaluates to:\[W = 2 \int_{0}^{60} (100 - \frac{1}{4}t) \, dt = 2 \left[100t - \frac{1}{8}t^2 \right]_{0}^{60}\]\[W = 2 \left[100(60) - \frac{1}{8}(60)^2 \right] = 2 \left[6000 - 450 \right] = 2(5550) = 11100 \, ext{foot-pounds}\]

Conclusion

The total work done in lifting the bag, accounting for the leaking sand over the uniform height of 120 ft, is 11100 foot-pounds.

Key concepts

Variable Force Integration

When dealing with problems where the force applied is not constant, we use the method of variable force integration to find work done. In this problem, the sand bag's weight decreases as sand leaks out, causing the force required to lift it to change continuously.

To solve it, we integrate the variable force over the distance the bag is lifted. At any time \( t \), the force exerted is the weight of the bag at that moment. Since sand leaks at \( \frac{1}{4} \mathrm{lb/s} \), the force at time \( t \) is \( 100 - \frac{1}{4}t \) pounds. To find total work, we integrate this force over the lifting distance.

Calculus Applications

Calculus is a powerful tool used in solving real-life problems involving continuous change like here, where the sand leaks out over time as the bag is lifted. This problem involves applying calculus to find the work done as an integral.

• We represent the variable force with a function: \( F(t) = 100 - \frac{1}{4}t \).
• The distance function for the lifting process is \( x(t) = 2t \), indicating that the distance covered after \( t \) seconds is \( 2t \) feet.

We use these functional relationships to set up the integral and calculate the work done.

Weight Loss Rate

The rate at which the sand is leaking forms an important part of the problem and affects how we calculate the changing force. In this scenario, the bag loses sand at a rate of \( \frac{1}{4} \mathrm{lb/s} \).

Over a period of 60 seconds, the total weight loss sums up to \( 15 \mathrm{lb} \). This means initially the bag weighs \( 100 \mathrm{lb} \) and eventually weighs \( 85 \mathrm{lb} \) when all sand loss is calculated. Understanding the weight loss rate helps determine the decreasing force on the bag, which is essential for successful integration and calculating overall work.

Distance Integration

Integration over a distance refers to finding the total work done when a force moves an object over a set distance, while the force varies. Here, distance integration involves integrating the decreasing force function over the path of 120 feet.

The integration problem is set up by expressing the force in terms of time and substituting it into the integral for work over distance:\[W = \int_0^{60} F(t) \cdot dx = \int_0^{60} (100 - \frac{1}{4}t) \cdot 2 \, dt\]This approach, calculating the integral of force with respect to the moved distance, enables us to calculate the total work done when lifting the sandbag even as its weight changes.