Problem 39 - University Physics with Modern Physics

Question

Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on the frictionless surface of a frozen pond, connected by a light rope (\(\textbf{Fig. P4.39}\)). A woman wearing golf shoes (for traction) pulls horizontally on the 6.00-kg crate with a force \(F\) that gives the crate an acceleration of 2.50 m/s\(^2\). (a) What is the acceleration of the 4.00-kg crate? (b) Draw a free-body diagram for the 4.00-kg crate. Use that diagram and Newton's second law to find the tension \(T\) in the rope that connects the two crates. (c) Draw a free-body diagram for the 6.00-kg crate. What is the direction of the net force on the 6.00-kg crate? Which is larger in magnitude, \(T\) or \(F\)? (d) Use part (c) and Newton's second law to calculate the magnitude of \(F\).

Step-by-Step Solution

Answer

(a) 2.50 m/s²; (b) T = 10.0 N; (c) Forward; (d) F = 25.0 N.

Step 1: Understand the Problem

We are given two crates connected by a rope on a frictionless surface. The problem provides the mass of each crate and the acceleration of the 6.00-kg crate. We need to find the acceleration of the 4.00-kg crate, the tension in the rope, and the magnitude of the pulling force.

Step 2: Analyze the 4.00-kg Crate

Since there is no friction, both crates will have the same horizontal acceleration. Thus, the acceleration of the 4.00-kg crate is also 2.50 m/s².

Step 3: Draw Free-Body Diagram for 4.00-kg Crate

The only horizontal force acting on the 4.00-kg crate is the tension in the rope, labeled as T.

Step 4: Apply Newton's Second Law to 4.00-kg Crate

Using Newton's second law, calculate the tension in the rope: \( T = m_2 \cdot a \), where \( m_2 = 4.00 \; \text{kg} \) and \( a = 2.50 \; \text{m/s}^2 \). Thus, \( T = 4.00 \cdot 2.50 = 10.0 \; \text{N} \).

Step 5: Draw Free-Body Diagram for 6.00-kg Crate

For the 6.00-kg crate, there are two forces: tension \( T \) pointing backward, and the force \( F \) pulling forward.

Step 6: Determine Net Force on 6.00-kg Crate

The net force \( F_{ ext{net}} \) acting on the 6.00-kg crate is the forward force \( F \) minus the tension \( T \). Both forces act horizontally.

Step 7: Compare Magnitudes of Tension and Force F

Since the 6.00-kg crate accelerates forward, \( F \) must be greater than \( T \).

Step 8: Use Newton's Second Law for 6.00-kg Crate to Find F

Using \( F - T = m_1 \cdot a \), we have \( F = T + m_1 \cdot a \), where \( m_1 = 6.00 \; \text{kg} \), \( a = 2.50 \; \text{m/s}^2 \), and \( T = 10.0 \; \text{N} \). Calculating \( F \) gives \( F = 10.0 + 6.00 \cdot 2.50 = 25.0 \; \text{N} \).

Key Concepts

Frictionless Surface

In the problem, we have a frictionless surface which means there is no frictional force opposing the movement of the crates. This concept is crucial because it simplifies the motion analysis. Since there is no friction, the net force acting on the blocks is directly the result of the applied forces.
When analyzing problems involving frictionless surfaces, we can make a few assumptions:

• No energy is lost due to friction, meaning all forces are either applied forces or tensions.
• Both objects in contact or connected move together if the force is applied horizontally.
• It's easier to calculate net forces and analyze motion since we only focus on the applied forces, minimizing unnecessary complexities.

Thus, on a frictionless surface, any applied force results in acceleration, as it's not countered by resistance from friction.

Tension in Physics

Tension acts along the rope that connects the two crates, which is important in transferring force between them. Tension is essentially a pulling force exerted by a string, cable, or rope on an object, directed along the length of the medium.
Here's what we should understand about tension in this context:

• Tension magnitude should be the same throughout the rope as long as it's light and assumed to be ideal, meaning it doesn't stretch and has negligible mass.
• In the exercise, tension transmits the pulling force from the woman to the 4.00-kg crate, allowing it to accelerate.
• We calculated this tension using Newton's second law for the 4.00-kg crate, finding it to be 10.0 N by considering its mass and acceleration.

Understanding tension helps in analyzing how forces are transmitted through mediums linking different components in a system.

Free-Body Diagram

A free-body diagram helps visualize forces acting on an object, crucial for solving physics problems effectively. It's a sketch that represents a single object and the applied forces.
Let's break down its elements:

• Each force acting on the object is represented by an arrow pointing in its direction; the length reflects the magnitude.
• For the 4.00-kg crate, the diagram includes only one horizontal force: the tension in the rope.
• For the 6.00-kg crate, it illustrates two forces: the pulling force (\(F\)) to the right and tension (\(T\)) to the left.

Free-body diagrams simplify complex systems, making it easier to apply Newton’s laws, calculate net forces, and understand the physical dynamics of the system.

Net Force

The net force is the total force acting on an object after all opposing forces are accounted for. According to Newton's second law, it's given by the equation \(F_{\text{net}} = ma\), where \(m\) represents mass, and \(a\) is acceleration.
Here's a breakdown of the concept in our exercise:

• For the 6.00-kg crate, the net force is calculated as the difference between the applied force (\(F\)) and the tension (\(T\)).
• The positive direction is generally chosen in the direction of acceleration, hence \(F - T\) equals the product of mass and acceleration.
• Concerning the larger force, \(F\) must exceed \(T\), allowing the crate to accelerate forward.

Understanding net force ensures a clear analysis of how unopposed force results in acceleration, crucial for solving dynamics problems.