Question

A typical textbook may have a mass of \(1 \mathrm{~kg}\), and thus a weight of about \(10 \mathrm{~N}\). How high could the textbook be lifted (against the force of gravity) by supplying one Joule of energy?

Short answer

Answer: 0.1 meters (10 centimeters)

Step-by-step solution

Identify the formula for work done

Work done (W) is the energy transferred when a force (F) acts on an object over a distance (d). The formula for work done is: W = F * d * cos(theta) In this case, we are lifting the textbook against the force of gravity, which means the angle (theta) between the force and the distance is 0 degrees and cos(0) = 1. So, the formula simplifies to: W = F * d

Rearrange the formula to solve for distance

We are given the values for W and F, and we need to find the value of d. We can rearrange the formula for work done to solve for distance (d): d = W / F

Substitute given values and solve for distance

Now substitute the given values for W (1 Joule) and F (10 N) into the formula and solve for distance (d): d = (1 J) / (10 N) d = 0.1 m

Interpret the result

The textbook could be lifted 0.1 meters (10 centimeters) high when supplied with one Joule of energy.

Key concepts

Work and Energy

Understanding work and energy is fundamental to solving physics problems, specifically when dealing with forces and motion. Work is done when a force causes an object to move in the direction of the force applied. The equation representing work is expressed as:

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

Here, \( W \) is the work done or energy transferred, \( F \) is the force applied, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and direction of movement. When lifting an object against gravity, \( \theta = 0 \) degrees, making \( \cos(0) = 1 \), simplifying the equation to \( W = F \times d \). This formula helps us find out how much energy is required to move an object over a given distance.

In the context of our textbook problem, the fact that the textbook is being lifted straight up makes calculating the work done straightforward since the angle \( \theta \) does not cause any change in direction.

Gravitational Force

Gravitational force is a force that attracts any two objects with mass. It is what gives weight to physical objects and is experienced as the force that pulls everything towards the center of the Earth. The force due to gravity is given by the formula:

• \( F_g = m \times g \)

Where \( F_g \) is the gravitational force, \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, approximately \( 9.8 \mathrm{~m/s^2} \) on Earth's surface.

In our problem, the textbook has a mass of 1 kg, making the gravitational force \( 10 \mathrm{~N} \) using \( F_g = 1 \mathrm{~kg} \times 9.8 \mathrm{~m/s^2} \approx 10 \mathrm{~N} \). This indicates that to lift the textbook, the force applied must counteract this gravitational force.

Textbook Lifting Problem

This problem helps to understand the practical application of work and energy concepts, specifically how much energy is required to lift an object against gravitational pull. Here, the goal is to determine how high we can lift a textbook by supplying a certain amount of energy, specifically 1 Joule.

The problem solution involves finding the height reach using the rearranged work formula:

• \( d = \frac{W}{F} \)

Given \( W = 1 \mathrm{~J} \) and \( F = 10 \mathrm{~N} \), we calculate the distance \( d \) as:

• \( d = \frac{1 \mathrm{~J}}{10 \mathrm{~N}} = 0.1 \mathrm{~m} \)

This means that with 1 Joule of energy, you can lift the textbook 0.1 meters, or 10 centimeters, off the ground. Understanding this small height illustrates how gravitational force impacts energy consumption. It shows why lifting heavier objects requires significantly more energy.