Question

Two blocks of equal mass are connected by a massless horizontal rope and resting on a frictionless table. When one of the blocks is pulled away by a horizontal external force \(\vec{F}\) what is the ratio of the net forces acting on the blocks? a) 1: 1 c) 1: 2 b) 1: 1.41 d) none of the above

Short answer

Answer: d) none of the above. The ratio of the net forces acting on the two blocks is F - T, where F is the external force and T is the tension force in the rope.

Step-by-step solution

Analyze the forces acting on Block 1

Let's focus on the block that has the external force \(\vec{F}\) applied to it (let's call it Block 1). The forces acting on Block 1 are the external force \(\vec{F}\) and the tension force \(\vec{T}\) due to the rope. Since both forces are in the horizontal direction and opposing each other, we can write Newton's second law for Block 1 as: \(F - T = m_1a_1\) (1)

Analyze the forces acting on Block 2

Now let's focus on the other block (let's call it Block 2). The only force acting on it is the tension force \(\vec{T}\) due to the rope. We can write Newton's second law for Block 2 as: \(T = m_2a_2\) (2)

Express both accelerations in terms of tension and masses

From equation (1), we can express \(a_1\) in terms of tension and the mass of Block 1: \(a_1 = \frac{F - T}{m_1}\) (3) From equation (2), we can express \(a_2\) in terms of tension and the mass of Block 2: \(a_2 = \frac{T}{m_2}\) (4)

Find the ratio of accelerations

Divide equation (4) by equation (3) to find the ratio of \(a_2\) to \(a_1\): \(\frac{a_2}{a_1} = \frac{\frac{T}{m_2}}{\frac{F - T}{m_1}}\) (5) Since both blocks have equal masses (as stated in the exercise), \(m_1 = m_2\), and equation (5) simplifies to: \(\frac{a_2}{a_1} = \frac{T}{F - T}\) (6)

Find the ratio of net forces

To find the ratio of the net forces acting on the two blocks, we need to relate their accelerations to the net forces. From Newton's second law, \(F_{net} = ma\), where \(F_{net}\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. Therefore, we have: \(\frac{F_{net1}}{F_{net2}} = \frac{m_1a_1}{m_2a_2}\) (7) Using equation (6) and the fact that both blocks have equal masses, we find: \(\frac{F_{net1}}{F_{net2}} = \frac{m_1a_1}{m_2\frac{T}{F - T}} = \frac{1}{\frac{1}{F - T}} = F - T\) (8)

Determine the ratio from the options given

From equation (8), we conclude that the ratio of the net forces acting on the two blocks is \(F - T\). Since \(T\) is always less than \(F\), the ratio cannot be equal to any of the options given (a, b, or c). Therefore, the answer is: d) none of the above

Key concepts

net force

When discussing Newton's Second Law, the concept of the net force is essential. It refers to the total force acting upon an object, calculated by summing all individual forces. According to the law, if you apply a force to an object, it will result in the object's mass being multiplied by its acceleration:

• The formula is given by: \( F_{net} = ma \)
• In our problem with the two blocks, Block 1 is acted upon by both the external force \( \vec{F} \) and the tension \( \vec{T} \), whereas Block 2 feels only the tension force from the rope.
• Since there is no friction, the net force acting on each block is directly connected to the acceleration we observe.

Understanding the net force enables us to predict how each block will move on the frictionless surface. Each block moves according to the forces exerted on it, which can be additive like in Block 1's case, or singular as in Block 2's case.

tension

Tension plays a crucial role when two objects are connected by a rope or a similar medium. It is the force transmitted through the rope, pulling equally on the objects it connects.

• In this exercise, tension \( \vec{T} \) is the only force acting on Block 2, stabilizing and pulling it along the table.
• The tension in the rope contributes to the net force acting on Block 1 but opposes the external force \( \vec{F} \).

The balance this sets up ensures both blocks move together. The tension adjusts accordingly when either block experiences a force, maintaining a consistent and unified acceleration among them.

acceleration

Acceleration is a fundamental aspect of motion resulting from net forces. In this exercise, both blocks experience acceleration due to forces acting on them.

• Block 1's acceleration can be expressed by \( a_1 = \frac{F - T}{m_1} \), demonstrating the combined impact of the external force and tension.
• Block 2's motion is controlled solely by the tension, characterized by \( a_2 = \frac{T}{m_2} \).

Owing to the equal mass of the blocks, the accelerations tend to synchronize, particularly when considering the blocks' identical mass and interconnected behavior via the rope. Observing these accelerations helps breakdown the motion dynamics of such linked systems.

equal mass blocks

The feature of having equal mass in a system simplifies the analysis due to identical gravitational and inertial properties.

• When masses are equal, as they are here, \( m_1 = m_2 \), it provides a base for simplifying expressions like \( \frac{a_2}{a_1} = \frac{T}{F - T} \).
• This equality implies a direct correlation between the acceleration of each block, so calculations of forces become consistent and predictable.

Having blocks of identical mass is a beneficial scenario in physics problems as it allows us to focus on other variables like tension and external forces. The blocks will mirror each other's responses, offering straightforward, symmetrical solutions to force and motion problems.