Problem 55 - University Physics with Modern Physics

Question

Working Like a Horse. Your job is to lift 30 -kg crates a vertical distance of 0.90 m from the ground onto the bed of a truck. (a) How many crates would you have to load onto the truck in 1 minute for the average power output you use to lift the crates to equal 0.50 \(\mathrm{hp} ?\) (b) How many crates for an average power output of 100 \(\mathrm{W} ?\)

Step-by-Step Solution

Answer

(a) 85 crates in 1 minute for 0.50 hp. (b) 23 crates for 100 W.

Step 1: Understand the Problem

The exercise is about calculating the number of crates needed to achieve a specific power output when lifting them to a certain height. For part (a), the power output is 0.50 horsepower (hp), and for part (b), it is 100 watts (W).

Step 1: Convert Power from Horsepower to Watts

One horsepower is equivalent to 746 watts. Therefore, the power output of 0.50 hp is converted as follows:\[0.50 \text{ hp} \times 746 \frac{\text{W}}{\text{hp}} = 373 \text{ W}\].

Step 2: Calculate Work Done per Crate

The work done in lifting one crate is equal to the gravitational potential energy gained by the crate, given by the formula:\[W = m \cdot g \cdot h\]where \(m = 30 \text{ kg}\), \(g = 9.8 \text{ m/s}^2\) (acceleration due to gravity), and \(h = 0.90 \text{ m}\). Substituting the values:\[W = 30 \cdot 9.8 \cdot 0.90 = 264.6 \text{ J}\].

Step 3: Calculate Number of Crates for 0.50 hp

To find the number of crates \(n\), use the formula:\[\text{Power} = \frac{\text{Total Work}}{\text{Time}} = \frac{n \cdot W}{t}\]where power is 373 W, \(t = 60\) seconds (1 minute), and \(W = 264.6 \text{ J}\). Rearrange and solve for \(n\):\[n = \frac{373 \times 60}{264.6} \approx 84.7\].Round up since you can't have a fraction of a crate, so \(n = 85\) crates.

Step 4: Calculate Number of Crates for 100 W

Repeat the same calculation for 100 W power:\[n = \frac{100 \times 60}{264.6} \approx 22.7\].Round up, so \(n = 23\) crates.

Key Concepts

Work and Energy

Work and energy are central concepts in physics, especially when discussing power calculation problems like the one involving lifting crates. Work is done when a force acts on an object to move it through a distance. The work done, often denoted as \(W\), is calculated by the formula:

• \(W = F \cdot d \cdot \cos(\theta)\)

where:

• \(F\) is the force applied,
• \(d\) is the distance moved,
• \(\theta\) is the angle between the force and the direction of motion.

If the motion is vertical, like lifting crates, the angle \(\theta\) is 0 degrees, making the \(\cos(0) = 1\). Hence, work done is simply the force times the distance. In our problem, the force is the weight of the crate, which is equal to its mass \(m\) times gravitational acceleration \(g\). This gives us the work done per crate:

• \(W = m \times g \times h\).

This formula highlights the link between work and energy, specifically gravitational potential energy, as the work done in lifting the crate is converted to potential energy.

Horsepower to Watts Conversion

Understanding how to convert horsepower to watts is essential when working with power calculations in physics, particularly dealing with motor or engine power. Horsepower (hp) is a unit of power that originated in the 18th century when horses were the main source of machinery.
To match modern standards, horsepower is often converted to watts. One horsepower is defined as 746 watts. This means:

• 1 hp = 746 W.

In the exercise problem, the power provided by 0.50 horsepower needs to be expressed in watts to calculate the number of crates lifted. Therefore, we multiply 0.50 hp by 746 W/hp to get:

• 0.50 hp \(\times\) 746 \(= 373\) W.

Converting power into common units like watts allows for uniform calculations, ensuring that different forms of power measurements can be compared or used within the same problem effortlessly.

Gravitational Potential Energy

Gravitational potential energy is a type of energy that an object possesses due to its position in a gravitational field, commonly experienced as the Earth's gravity. The potential energy \(U\) for an object of mass \(m\) lifted to a height \(h\) can be calculated using the equation:

• \(U = m \cdot g \cdot h\)

where:

• \(m\) is the mass of the object,
• \(g\) is the acceleration due to gravity (approximately \(9.8 \ \text{m/s}^2\) on Earth),
• \(h\) is the height above the reference point.

In the exercise, lifting a 30 kg crate over a height of 0.90 meters on Earth requires calculating the potential energy (or the work done since energy is transferred) using the given formula. This results in:

• \(U = 30 \times 9.8 \times 0.9 = 264.6 \ \text{J}\)

Understanding how potential energy works in this context helps illustrate why it's necessary to account for the energy involved in lifting objects, whether for moving crates or assessing any lifting task.