Problem 1 - University Physics with Modern Physics

Question

You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-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

3.60 J by push, -0.900 J by friction, 0 J by normal force, 0 J by gravity, and 2.70 J net work.

Step 1: Calculate Work Done by Your Push

The work done by a force is calculated using the formula: \( W = F \times d \times \cos(\theta) \), where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion. For your 2.40-N push, the force is in the same direction as the motion, so \( \theta = 0^\circ \) and \( \cos(0^\circ) = 1 \). Thus, \( W_{push} = 2.40 \times 1.50 \times 1 = 3.60 \, \text{J} \).

Step 2: Calculate Work Done by Friction

The force of friction opposes the direction of motion, so \( \theta = 180^\circ \) and \( \cos(180^\circ) = -1 \). The work done by friction is \( W_{friction} = 0.600 \times 1.50 \times (-1) = -0.900 \, \text{J} \).

Step 3: Calculate Work Done by the Normal Force

The normal force acts perpendicular to the direction of motion. Since \( \theta = 90^\circ \) for the normal force and \( \cos(90^\circ) = 0 \), the work done by the normal force is \( W_{normal} = 0 \).

Step 4: Calculate Work Done by Gravity

Gravity also acts perpendicular to the horizontal direction of motion. Since \( \theta = 90^\circ \) for gravity and \( \cos(90^\circ) = 0 \), the work done by the force of gravity is \( W_{gravity} = 0 \).

Step 5: Calculate the Net Work Done on the Book

The net work done on the book is the sum of the work done by all forces. Thus, \( W_{net} = W_{push} + W_{friction} + W_{normal} + W_{gravity} = 3.60 + (-0.900) + 0 + 0 = 2.70 \, \text{J} \).

Key Concepts

Understanding Force

Force is a push or pull exerted on an object. It can change the object's speed, direction, or shape. In our exercise, there are several forces at play. These include the push you exert on the book and the opposing frictional force.
Forces are vector quantities, meaning they have both magnitude and direction. When calculating work done by these forces, it's important to know the angle between the force direction and the motion. This angle helps determine the component of the force that contributes to moving the object, as seen in the equation for work:

• The formula for work is: \( W = F \times d \times \cos(\theta) \)

In this case, your push is fully aligned with the motion (\( \theta = 0^\circ \)), making all its magnitude effective for doing work.

The Role of Friction

Friction is a force that opposes the motion of objects sliding against each other. In our scenario, it acts to resist the movement of the book as you push it. Friction is crucial because it prevents objects from sliding indefinitely when there's no continuous external push.
When calculating work done by friction, we note that it acts in the opposite direction to the push. This means the angle \( \theta \) in the work formula becomes \( 180^\circ \). Hence, the work done by friction is negative (\( W_{friction} = -0.900 \, \text{J} \)), indicating that it takes energy away from the system.

• Frictional work can be a vital consideration in energy budgets, where losses are significant.

Normal Force Explained

The normal force acts perpendicular to the surface in contact with an object. It is a reactive force that balances the weight of an object resting on a surface, preventing it from falling through.
In the exercise, the normal force comes from the tabletop and acts upward, perpendicular to the horizontal push. Since it doesn't have a component in the direction of motion (\( \theta = 90^\circ \)), it does no work on the book:

• The work done is zero since \( \cos(90^\circ) = 0 \).

The normal force is crucial for determining how other forces act on objects, often appearing in calculations involving friction.

Gravity's Role

Gravity is the force that attracts objects toward the Earth. It acts vertically downward, and although prominent generally, it doesn't contribute to work in horizontal movements like our book's slide.
In this case, gravity acts at \( \theta = 90^\circ \) to the direction of motion, so it doesn't impact the work done on the book. Like the normal force, it does zero work here:

• Gravity's fundamental impact is in vertical motion, providing energy for movement such as falling.

Understanding gravity helps in analyzing energy transitions and movements in physics.

Calculating Net Work

Net work is the total work done considering all forces acting on an object. It reveals how much energy is transferred due to these forces, accounting for opposing and supporting influences.
To calculate net work, sum up the work from all individual forces. Here, it's the combination of your push, friction, the normal force, and gravity:

• \( W_{net} = 3.60 \text{ J} + (-0.900 \text{ J}) + 0 \text{ J} + 0 \text{ J} = 2.70 \text{ J} \)
• Net work helps determine the resultant energy change in the object's motion.

Net work indicates whether an object accelerates, tends to move, or remains stationary based on total energy dynamics.