Problem 6 - Applied Physics

Question

When a \(36 \overline{0} 0\) -lb automobile runs out of gas, it is pushed by its unhappy driver and a friend a quarter of a mile \((0.250 \mathrm{mi})\). To keep the car rolling, they must exert a constant force of \(175 \mathrm{lb}\). (a) How much work do they do? (b) If it takes them \(15.0\) min, how much power do they develop? (c) Expressed in horsepower, how much power do they develop?

Step-by-Step Solution

Answer

(a) 231,000 ft-lb, (b) 256.67 ft-lb/s, (c) 0.4676 hp

Step 1: Understanding the problem and identifying given values

To solve this problem, we need to determine the work done by the force applied to the car, the power developed, and express the power in horsepower. The given values are: - Constant force exerted: 175 lb - Distance over which the force is exerted: 0.250 mi - Time taken: 15.0 minutes.

Step 2: Convert distance from miles to feet

Since the force is given in pounds and we usually measure work in foot-pounds in this context, convert the distance from miles to feet. We know that 1 mile = 5280 feet. Therefore, \[0.250 \text{ miles} = 0.250 \times 5280 \text{ feet} = 1320 \text{ feet}\]

Step 3: Calculate the work done

The work done is given by the formula \(W = F \times d\), where \(F\) is the force and \(d\) is the distance. Both values are now in compatible units (pounds and feet, respectively). So, the work done is:\[W = 175 \text{ lb} \times 1320 \text{ ft} = 231000 \text{ ft-lb}\]

Step 4: Convert time from minutes to seconds

To find the power, we need to convert time from minutes to seconds. We know that 1 minute = 60 seconds:\[15.0 \text{ minutes} = 15.0 \times 60 = 900 \text{ seconds}\]

Step 5: Calculate power in foot-pounds per second

Power is defined as work done per unit of time. Thus, calculate power using the formula:\[P = \frac{W}{t}\]where \(W\) is the work, and \(t\) is the time in seconds:\[P = \frac{231000 \text{ ft-lb}}{900 \text{ s}} = 256.67 \text{ ft-lb/s}\]

Step 6: Convert power to horsepower

Horsepower is a unit of power where 1 horsepower is equivalent to 550 ft-lb/s. To convert the power we calculated to horsepower, use the formula:\[\text{Power in horsepower} = \frac{256.67 \text{ ft-lb/s}}{550 \text{ ft-lb/s per horsepower}}\]This gives us:\[\text{Power in horsepower} = 0.4676 \text{ hp}\]

Key Concepts

Converting Units in Physics

In physics, working with compatible units is crucial for solving problems accurately. In our exercise, we started with measurements given in different units: force in pounds, distance in miles, and time in minutes.
To ensure these measurements are compatible, we first convert:

• Distance: Since force is measured in pounds, and work involves force times distance, it was necessary to convert miles to feet. The conversion is straightforward, using the known relation: 1 mile = 5280 feet. Thus, 0.250 miles equals 1320 feet.
• Time: Since power calculations often require time in seconds, convert 15 minutes into seconds. With 1 minute equal to 60 seconds, 15 minutes becomes 900 seconds.

Conversions ensure that each part of the formula uses consistent units, facilitating correct calculation of work and power.

Mechanics in Physics

Mechanics is a branch of physics dealing with motion and forces. The original exercise involves concepts of force, distance, work, and power—all foundational to mechanics. The primary formula used here is for work, defined as the product of force and distance moved by the object:\[ W = F \times d \]Where:

• \( W \) is the work done.
• \( F \) is the force applied, here 175 pounds.
• \( d \) is the distance covered, 1320 feet after conversion.

Understanding this relationship is vital in mechanics. The force must act consistently over a distance, which directly impacts the work done on an object. This concept broadly applies to many other real-world scenarios.

Automobile Physics Problem

This exercise is a practical example within automobile physics. When an automobile runs out of fuel and must be pushed, the amount of work involved depends significantly on the force exerted and the distance moved. This problem helps explore:

• How forces interact with an object, here the car, and how the object is moved across a distance due to these forces.
• The real-world application of physics principles, showing how work and energy must be expended even for everyday tasks like moving a car.

Applying principles of work and power in the context of an automobile provides insight into how vehicles consume energy, helping students understand both simple and complex mechanical operations.

Power in Horsepower

Power relates to how quickly work is done. In this problem, it calculates how fast the friends exerted force to move the car over a certain period, using:\[ P = \frac{W}{t} \]Where:

• \( P \) is the power.
• \( W \) is the work done, 231000 ft-lb.
• \( t \) is time, 900 seconds.

Calculating power in foot-pounds per second gives insight into efficiency. To convert this power into more familiar terms, horsepower – a unit often associated with engines and machinery – is used.

• 1 horsepower equals 550 foot-pounds per second.
• By dividing the calculated power by 550, we find power in horsepower units: 0.4676 hp.

Conversion into horsepower can be particularly relevant in automotive contexts, offering a tangible understanding of how energy outputs relate to engine performance.