Problem 1 - University Physics with Modern Physics

Question

You push your physics book 1.50 \(\mathrm{m}\) along a horizontal table top with a horizontal push of 2.40 \(\mathrm{N}\) while the opposing force of friction is 0.600 \(\mathrm{N} .\) How much work does each of the following forces do on the book: (a) your \(2.40-\mathrm{N}\) push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

Step-by-Step Solution

Answer

(a) 3.60 J; (b) -0.90 J; (c) 0 J; (d) 0 J; (e) 2.70 J total net work.

Step 1: Understand the Work Formula

The work done by a force is calculated using the formula \( W = F \times d \times \cos(\theta) \), where \( F \) is the magnitude of the force, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of movement.

Step 2: Calculate Work Done by Push Force

For the force of your push, \( F = 2.40 \, \mathrm{N} \), \( d = 1.50 \, \mathrm{m} \), and \( \theta = 0^\circ \) since the push is horizontal and in the direction of movement. The work done is \( W = 2.40 \times 1.50 \times \cos(0) = 3.60 \, \mathrm{J} \).

Step 3: Calculate Work Done by Friction

The friction force opposes movement, so \( F = 0.600 \, \mathrm{N} \), \( d = 1.50 \, \mathrm{m} \), and \( \theta = 180^\circ \). The work done is \( W = 0.600 \times 1.50 \times \cos(180) = -0.90 \, \mathrm{J} \).

Step 4: Calculate Work Done by Normal Force

The normal force acts perpendicular to the displacement, so \( \theta = 90^\circ \), making the work done \( W = F \times d \times \cos(90) = 0 \) because \( \cos(90) = 0 \).

Step 5: Calculate Work Done by Gravity

Similarly, the gravitational force also acts perpendicular to the displacement. Thus, the angle \( \theta = 90^\circ \) and \( W = F \times d \times \cos(90) = 0 \).

Step 6: Determine the Net Work Done

The net work done is the sum of the work done by all forces: Work by push + Work by friction + Work by normal force + Work by gravity = \( 3.60 + (-0.90) + 0 + 0 = 2.70 \, \mathrm{J} \).

Key Concepts

friction in physics

Friction is a crucial force in physics that resists the relative motion between two surfaces in contact. It can be your ally or your adversary, depending on the situation. It's why we can walk without slipping, yet it also slows down objects.

To better understand friction, remember these key points:

• **Types of Friction:** There are mainly two types: static friction, which prevents motion, and kinetic friction, which opposes ongoing motion.
• **Direction:** Friction always acts in the opposite direction to the applied force or movement. In our exercise, friction opposes your push on the book.
• **Magnitude:** The strength of friction depends on two main factors: the nature of the surfaces in contact and the normal force pressing them together.

In our scenario, the frictional force is 0.600 N, acting opposite to your push, and does negative work. This is typical because friction removes energy from the system, slowed down the motion of the book, and causes wear and tear.

net work done

The concept of net work done is about understanding the total energy change in an object as forces act upon it. Consider it as a balance of work contributed by each individual force.

In the given problem, we use the formula for work: \[ W = F \times d \times \cos(\theta) \] where \( F \) is force, \( d \) is distance, and \( \theta \) is the angle between force and direction of motion.

• **Work by Applied Force:** Here, the push does positive work, calculated as 3.60 J. This adds energy to the system, moving the book forward.
• **Negative Work of Friction:** Friction does -0.90 J of work, counteracting the push by taking away energy.
• **Zero Work by Perpendicular Forces:** The work by normal force and gravity is zero because they act perpendicular to the motion, and \( \cos(90^{\circ}) = 0 \).

Adding these contributions gives the net work done: 2.70 J. This shows the book had a net gain of energy enabling further motion after accounting for friction.

force and motion

Understanding the relationship between force and motion is key to mastering physics. An unbalanced force will cause an object to accelerate, which is central to Newton's laws of motion.

In our scenario, several forces interact to influence the motion of the book:

• **Applied Force:** The push you provide acts in the direction of the book's motion, causing it to move horizontally.
• **Frictional Force:** This opposes the applied force and aims to stop the book, representing resistance in the system.
• **Normal Force and Gravity:** These forces act vertically, balancing each other and not affecting horizontal motion. They ensure the book doesn't float away or fall through the table.

The lesson here is that motion results from unbalanced forces. The book moves because the push is stronger than the frictional force, leading to a positive net work done. This is a vivid demonstration of how forces and motion interplay in practical physics.