Problem 8
Question
A hay wagon is used to move bales from the field to the barn. The tractor pulling the wagon exerts a constant force of \(350 \mathrm{lb}\). The distance from field to barn is \(\frac{1}{2}\) mi. How much work (ft \(l\) b) is done in moving one load of hay to the barn?
Step-by-Step Solution
Verified Answer
The work done is 924,000 ft lb.
1Step 1: Convert Distance to Feet
First, convert the distance from miles to feet, because the work will be calculated in foot-pounds. Since there are 5280 feet in a mile:\[\text{Distance in feet} = \frac{1}{2} \times 5280 = 2640 \text{ feet}\]
2Step 2: Calculate Work Done
The work done (in foot-pounds) can be calculated using the formula:\[\text{Work} = \text{Force} \times \text{Distance}\]Substitute the given values:\[\text{Work} = 350 \text{ lb} \times 2640 \text{ feet} = 924,000 \text{ ft lb}\]
Key Concepts
Understanding Force in PhysicsConverting Distance in Physics ProblemsCalculating Work in Foot-PoundsApproaching Physics Problem Solving
Understanding Force in Physics
In physics, force is defined as any interaction that, when unopposed, will change the motion of an object. Force can cause an object to start moving, stop moving, or change direction. The unit for measuring force in the International System of Units (SI) is the Newton. However, in many physics problems, especially those using the Imperial system, we measure force in pounds (lb).
For example, in our hay wagon exercise, the tractor exerts a force of 350 pounds to pull the wagon across the field. This means that the tractor is applying a steady push or pull of 350 pounds against the wagon to keep it moving towards the barn.
Whenever you deal with force, remembering that it is a vector quantity is important. This means force has both magnitude and direction. In our problem, the magnitude is 350 lb, and the direction is towards the barn.
For example, in our hay wagon exercise, the tractor exerts a force of 350 pounds to pull the wagon across the field. This means that the tractor is applying a steady push or pull of 350 pounds against the wagon to keep it moving towards the barn.
Whenever you deal with force, remembering that it is a vector quantity is important. This means force has both magnitude and direction. In our problem, the magnitude is 350 lb, and the direction is towards the barn.
Converting Distance in Physics Problems
In many physics exercises, the need to convert units of distance arises to ensure consistency, especially when calculating work or energy. Work calculations often require the distance in feet when dealing with the Imperial system.
In our hay wagon scenario, the distance from the field to the barn is given as \(\frac{1}{2}\) mile. However, to calculate work in foot-pounds, you first need to convert this distance into feet.
This conversion is straightforward:
In our hay wagon scenario, the distance from the field to the barn is given as \(\frac{1}{2}\) mile. However, to calculate work in foot-pounds, you first need to convert this distance into feet.
This conversion is straightforward:
- There are 5280 feet in one mile.
- For half a mile, the distance in feet is \(\frac{1}{2} \times 5280 = 2640\) feet.
Calculating Work in Foot-Pounds
When we talk about work in physics, we're referring to an action that involves force moving an object over a distance. The formula used to calculate work is:
\[\text{Work} = \text{Force} \times \text{Distance}\]
In the context of the problem, work must be calculated in foot-pounds because both force and distance are measured in these units in the Imperial system.
\[\text{Work} = 350 \text{ lb} \times 2640 \text{ ft} = 924,000 \text{ ft lb}\]
This figure represents the total effort put into moving the hay from the field to the barn, expressed as work done in foot-pounds.
\[\text{Work} = \text{Force} \times \text{Distance}\]
In the context of the problem, work must be calculated in foot-pounds because both force and distance are measured in these units in the Imperial system.
- The force applied is 350 pounds (lb).
- The distance moved is 2640 feet, as calculated from the distance conversion.
\[\text{Work} = 350 \text{ lb} \times 2640 \text{ ft} = 924,000 \text{ ft lb}\]
This figure represents the total effort put into moving the hay from the field to the barn, expressed as work done in foot-pounds.
Approaching Physics Problem Solving
Problem-solving in physics involves a step-by-step process where understanding the given problem and breaking it into manageable parts is key.
Here's how you can approach solving such problems effectively:
Here's how you can approach solving such problems effectively:
- Identify the known values: Write down the values given in the problem, such as force or distance.
- Convert units when necessary: Ensure all measurements are in compatible units before performing calculations, like converting miles to feet.
- Apply the appropriate formulas: Use the correct formula to link the known values to what you need to find, such as the work formula \(\text{Work} = \text{Force} \times \text{Distance}\).
- Perform the calculations: Substitute the values into the formula and solve for the unknown.
Other exercises in this chapter
Problem 8
An ironworker drops a hammer \(5.25 \mathrm{~m}\) to the ground. What is its speed as it hits the ground?
View solution Problem 8
A crate of mass \(475 \mathrm{~kg}\) is raised to a height \(17.0 \mathrm{~m}\) above the floor. What potential energy has it acquired with respect to the floor
View solution Problem 9
A box is dropped \(3.60 \mathrm{~m}\) to the ground. What is its speed as it hits the ground?
View solution Problem 9
How long would it take a \(95 \overline{0}-\mathrm{W}\) motor to raise a \(36 \overline{0}-\mathrm{kg}\) mass to a height of \(16.0 \mathrm{~m} ?\)
View solution