Problem 3 - University Physics with Modern Physics

Question

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Step-by-Step Solution

Answer

(a) The worker must apply a force of 73.5 N. (b) Work done by the force is 330.75 J. (c) Work done by friction is -330.75 J. (d) Work done by normal force and gravity is 0 J each. (e) Total work done is 0 J.

Step 1: Understanding the Force Required

To keep the crate moving at a constant velocity, the worker must apply a force equal to the force of friction. The force of friction can be calculated using the formula: \( f = \mu \cdot N \). Here, \( \mu = 0.25 \) (coefficient of kinetic friction) and \( N = m \cdot g \). Given \( m = 30.0 \) kg and \( g = 9.8 \) m/s², \( N = 30.0 \times 9.8 = 294 \) N. Thus, \( f = 0.25 \times 294 = 73.5 \) N. Therefore, the worker must apply a force of 73.5 N.

Step 2: Calculating Work Done by Worker

Work done by the worker is given by \( W = F \cdot d \), where \( F \) is the force applied and \( d \) is the distance. Here \( F = 73.5 \) N and \( d = 4.5 \) m. Therefore, \( W = 73.5 \times 4.5 = 330.75 \) J.

Step 3: Calculating Work Done by Friction

Since the force applied by the worker equals the force of friction, the work done by friction is also \( W = f \cdot d = 73.5 \times 4.5 = -330.75 \) J. The work done by friction is negative because it is opposite to the direction of motion.

Step 4: Work Done by Normal Force and Gravity

The normal force and gravity do work perpendicular to the displacement, hence \( W = 0 \) for both. The normal force acts vertically upwards and gravity acts vertically downwards, both of which have no component in the direction of horizontal displacement.

Step 5: Calculating Total Work Done on Crate

The total work done on the crate is the sum of the work done by all forces. Since the work done by the applied force and friction cancels each other out, and the work done by gravity and the normal force is zero, the total work done is \( 0 \) J.

Key Concepts

Work Done

In physics, the concept of "work done" refers to the action of a force causing an object to move. It is an essential concept when solving dynamics problems like the one with the factory worker and the crate. The formula for calculating work done is:

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

Here:

• \( W \) is the work done in joules (J),
• \( F \) is the force applied in newtons (N),
• \( d \) is the displacement in meters (m), and
• \( \theta \) is the angle between the force and the direction of displacement.

For the crate moving across a flat surface, the force is applied horizontally, and the angle \( \theta = 0 \), hence, \( \cos(0) = 1 \). This simplifies the work done by the force to \( W = F \cdot d \). Understanding how work is calculated helps in evaluating energy changes in the system.

Force Calculation

Force calculation plays a vital role in physics problem-solving and understanding how objects interact. In the example where a worker pushes a crate, it's important to calculate the required force to overcome friction and keep the crate moving at a constant speed. The force of friction is calculated using:

• \( f = \mu \cdot N \)

Where:

• \( \mu \) is the coefficient of kinetic friction between the surfaces,
• \( N \) is the normal force, which, on a flat surface, equals the weight of the object \( (N = m \cdot g) \).

Given:

• The mass \( m = 30.0 \) kg, and gravitational force \( g = 9.8 \) m/s²,
• The normal force \( N = 30.0 \times 9.8 = 294 \) N.

So, the frictional force \( f = 0.25 \times 294 = 73.5 \) N. Thus, the worker needs to exert a force of 73.5 N to counterbalance the frictional force.

Newton's Laws of Motion

Newton's Laws of Motion are critical to understanding dynamics problems. The factory worker example heavily relies on these laws, particularly the first law, often referred to as the Law of Inertia. According to this law:

• An object at rest stays at rest and an object in motion stays in motion at a constant velocity unless acted upon by a net external force.

In this context, the worker's force is precisely balancing the force of kinetic friction, resulting in no net force and allowing the crate to move at a constant velocity. Only when there is a balance between these forces, the crate remains in uniform motion conditions. This illustrates how forces interact based on Newton's principles.

Physics Problem Solving

Effective physics problem-solving includes a step-by-step approach to dissecting and analyzing problems. For scenarios like the crate problem:

• Begin by identifying all forces acting on the object.
• Use relevant physics formulas to compute values like force and work.
• Assess whether forces work in alignment, opposition, or perpendicularly in relation to displacement.
• Sum up individual work contributions from each force to find total work done.

By systematically applying these practices, problems become manageable and solutions emerge clearly. Recognizing how each force contributes or opposes motion simplifies even complex problems into solvable segments.