91Ó°ÊÓ

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the masses of both blocks are \(3.0 \mathrm{~kg} .\) In Case \(\mathrm{B}\) the mass of block \(1,\) the block behind, is \(6.0 \mathrm{~kg},\) and the mass of block 2 is \(3.0 \mathrm{~kg} .\) No frictional force acts on block 1 in either Case \(A\) or Case B. However, a kinetic frictional force does act on block 2 in both cases and opposes the motion. (a) Identify the forces that contribute to the horizontal net force acting on block \(1 .\) (b) Identify the forces that contribute to the horizontal net force acting on block 2 . (c) In which case, if either, do the blocks push against each other with greater forces? Explain. (d) Are the blocks accelerating or decelerating, and in which case, if either, is the magnitude of the acceleration greater? Problem The magnitude of the kinetic frictional force acting on block 2 in the drawing is \(5.8 \mathrm{~N}\). For both Case \(\mathrm{A}\) and Case \(\mathrm{B}\) deterine (a) the magnitude of the forces with which the blocks push against each other and (b) the acceleration of the blocks. Check to see that your answers are consistent with your answers to the Concept Questions.

Step by step solution

01

Identify Forces on Block 1

In both cases, block 1 experiences no friction. The only horizontal force acting on block 1 is the force of interaction between block 1 and block 2. This force is exerted by block 2 on block 1.

02

Identify Forces on Block 2

In both cases, block 2 experiences a kinetic frictional force of 5.8 N opposing its motion. Additionally, block 2 experiences the force exerted by block 1 on block 2 as it pushes the block forward.

03

Analyze Forces in Case A

For Case A, both blocks have the same mass of 3.0 kg. Let's denote the force between the blocks as \( F \). Applying Newton's second law to block 2 (3.0 kg): \[ F - 5.8 = 3.0 imes a \]Also, applying Newton's second law to block 1 (3.0 kg) gives:\[ F = 3.0 imes a \]

04

Solve Equations for Case A

From the equation for block 1: \[ a = \frac{F}{3.0} \]Substitute into the equation for block 2:\[ F - 5.8 = 3.0 \times \frac{F}{3.0} \]Solving gives:\[ F = 5.8 \text{ N} \] and \[ a = \frac{5.8}{3.0} = 1.93 \text{ m/s}^2 \].

05

Analyze Forces in Case B

For Case B, block 1 has a mass of 6.0 kg and block 2 has a mass of 3.0 kg. Applying Newton’s second law again, for block 2:\[ F - 5.8 = 3.0 imes a \]And for block 1:\[ F = 6.0 imes a \]

06

Solve Equations for Case B

From the equation for block 1:\[ a = \frac{F}{6.0} \]Substitute into the equation for block 2:\[ F - 5.8 = 3.0 \times \frac{F}{6.0} \]Solving gives:\[ F = 11.6 \text{ N} \] and \[ a = \frac{11.6}{6.0} = 1.93 \text{ m/s}^2 \].

07

Evaluate the Relative Forces and Acceleration

In both cases, the blocks decelerate due to friction. The pushing force is greater in Case B (11.6 N) than in Case A (5.8 N). The acceleration has the same magnitude (1.93 m/s²) in both cases but is actually a deceleration since the force opposes the motion.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frictional Force

Frictional force is a resistance force that occurs when two surfaces slide past each other. It opposes the motion and acts in the opposite direction to the applied force, playing a crucial role in determining the movement of objects.

In the context of the problem, block 2 experiences a frictional force as it slides over a surface. This frictional force is specified to be 5.8 N and acts against the movement of block 2.

Frictional force is determined by two main factors:

• The nature of the surfaces in contact
• The normal force, which is the perpendicular force exerted by the surface

Friction can be divided into two types: static friction, which prevents motion, and kinetic friction, which we will examine next.

Kinetic Friction

Kinetic friction occurs when two surfaces are in motion relative to each other. Unlike static friction, which is generally stronger as it prevents motion, kinetic friction acts to slow down or resist the motion of objects as seen with block 2 in this exercise.

The kinetic frictional force that opposes the sliding of block 2 is given as 5.8 N. We calculate kinetic friction using the formula:

• \( f_k = \mu_k \cdot N \)

where \( \mu_k \) is the coefficient of kinetic friction, a constant that varies based on materials and conditions, and \( N \) is the normal force. In a horizontal setup like ours, the normal force is equal to the weight of block 2 which is a product of its mass and the gravitational pull (mass \( \times \) gravity).

Understanding kinetic friction helps us comprehend why block 2 gradually slows down. The force that slows down the movement is consistent across the scenarios described, leading to deceleration.

Acceleration

Acceleration involves the change in velocity of an object due to applied forces and plays a vital part in the dynamics of moving bodies. In our situation, both blocks initially move at a certain speed and then experience forces that impact their motion.

Newton's second law of motion, expressed as \( F = m \cdot a \), relates the net force acting on an object to its mass and resulting acceleration (or deceleration). In our problem, the net force interacting with the blocks comprises the force of block 2 pushing against block 1, minus the force of kinetic friction acting on block 2.

For block 2, we calculate acceleration by rearranging:

• \( a = \frac{F}{m} \)

where \( F \) is the net force and \( m \) is the mass of the object. However, because the force exerted by friction opposes the block's motion, it causes deceleration rather than acceleration. The calculations in both Case A and Case B ensure that the blocks decelerate, reflecting the impacts of kinetic friction in real-world movement dynamics.