Problem 10 - College Physics

Question

A crate is dragged \(3.0 \mathrm{~m}\) along a rough floor with a constant velocity by a worker applying a force of \(500 \mathrm{~N}\) to a rope at an angle of \(30^{\circ}\) to the horizontal. (a) How many forces are acting on the crate? (b) How much work does each of these forces do? (c) What is the total work done on the crate?

Step-by-Step Solution

Answer

(a) Four forces, (b) Applied force: ~1299 J; Friction: ~-1299 J; Others: 0 J, (c) Total work done: 0 J.

Step 1: Identify the Forces Acting on the Crate

The forces acting on the crate include the force applied by the worker via the rope, the gravitational force acting downwards, the normal force from the floor acting upwards, and the frictional force opposing the movement. Thus, there are four forces in total.

Step 2: Calculate Work Done by the Applied Force

The work done by the applied force is given by the formula: \[ W = F \cdot d \cdot \cos(\theta) \]where \( F = 500 \, \mathrm{N} \), \( d = 3.0 \, \mathrm{m} \), and \( \theta = 30^{\circ} \). Substituting the values, we get:\[ W = 500 \times 3.0 \times \cos(30^{\circ}) = 500 \times 3.0 \times 0.866 \approx 1299 \mathrm{~J} \]

Step 3: Calculate Work Done by Friction

Since the crate moves with constant velocity, the frictional force does an equal and opposite work to the applied force, which means the work done by friction is approximately \(-1299 \mathrm{~J}\).

Step 4: Calculate Work Done by Gravitational and Normal Forces

Both gravitational and normal forces act perpendicular to the direction of movement, thus they do no work on the crate. Work is calculated as \( W = F \cdot d \cdot \cos(90^{\circ}) = 0 \mathrm{~J} \).

Step 5: Calculate the Total Work Done on the Crate

Since the crate is moving with constant velocity, the net work done on it is zero. This is because the work done by the applied force is positive and equal in magnitude to the negative work done by friction, while the other forces do no work.

Key Concepts

Forces in Physics

When dealing with problems like dragging a crate across a floor, understanding the different forces in physics is crucial. There are several forces that can act on objects, and it's important to be able to identify them to solve physics exercises.

• Applied Force: This is the force exerted by a person or another object. In our scenario, it's the 500 N force applied by the worker.
• Gravitational Force: Also known as weight, this force acts downward towards the center of the Earth. It is calculated by multiplying the mass of the object by the acceleration due to gravity, approximately 9.8 m/s².
• Normal Force: This is the support force exerted by a surface, perpendicular to the object resting on it. It balances the gravitational force in this scenario.
• Frictional Force: The force that opposes the motion and works against the applied force.

Recognizing these forces helps in understanding the dynamics involved in the motion of objects, particularly in equilibrium situations where they might cancel each other out, leading to constant velocity.

Frictional Force

Frictional force is a significant concept when analyzing motion on rough surfaces. This force arises due to the interlocking of surface irregularities between the crate and the floor.

• Direction: Friction always acts in the opposite direction to the movement. It's why a person must apply a constant force to keep the crate moving at a constant velocity.
• Calculation: While not directly calculated in the problem, we know the frictional force must equal the component of the applied force parallel to the floor when the crate moves at constant velocity. This results from Newton's first law, wherein no net force means no net acceleration.
• Negative Work: Friction does negative work when it opposes movement. This means it takes away energy that might otherwise cause acceleration.

Understanding friction is essential as it plays a critical role in bringing balance in systems and is often responsible for energy conversions within mechanical problems.

Gravitational Force

Gravitational force is one of the most fundamental forces in physics, affecting every object with mass. In our crate problem, this force acts downwards, but doesn't contribute to work since it acts perpendicular to the direction of movement.

• Direction: Always acts towards the center of the Earth. For our crate, it acts downwards and is balanced by the normal force from the floor.
• Work Calculation: Since work is the dot product of force and displacement, and because gravity acts perpendicularly to the motion of the crate, the work done by gravitational force is zero. This is calculated via the equation: \[ W = F \cdot d \cdot \cos(90^\circ) = 0 \text{ J} \]
• Impact: Even though it does no work when perpendicular, it is crucial in maintaining balance and determining the normal force. This factor impacts how frictional forces are calculated as they often depend on the normal force.

Gravitational force is a consistent factor in physics problems, and understanding its role is key to mastering the mechanics involving bodies at rest or in motion.