A Block Initially At Rest Is Given A Quick Push

When a block at rest receives a sudden push, its journey unfolds as a fascinating interplay of forces, inertia, and motion. Understanding this seemingly simple scenario requires delving into the fundamental principles of physics, including Newton's laws of motion, friction, and energy considerations.

Understanding the Initial State

Before the push, the block exists in a state of equilibrium. This means that the net force acting on it is zero. The forces at play are:

• Gravitational Force (Weight): The Earth pulls the block downwards.
• Normal Force: The surface on which the block rests pushes upwards, counteracting the gravitational force.

Since these forces are balanced, the block remains at rest, obeying Newton's First Law of Motion (the law of inertia). This law states that an object at rest will stay at rest unless acted upon by an external force.

The Quick Push: Introducing the Impetus

The "quick push" introduces an external force, momentarily disrupting the equilibrium. This force, let's call it Fpush, acts on the block for a short duration, imparting momentum.

• Impulse: The product of the force and the time interval during which it acts is called impulse (J). Mathematically, J = Fpush * Δt, where Δt is the duration of the push.
• Momentum: The impulse directly translates into a change in the block's momentum (p). Momentum is a measure of mass in motion and is calculated as p = mv, where m is the mass of the block and v is its velocity.
• Newton's Second Law: The push exemplifies Newton's Second Law (F = ma), where F is the net force, m is the mass, and a is the acceleration. During the push, the net force is primarily Fpush, causing the block to accelerate.

The key here is that the "quick push" doesn't sustain the force for an extended period. It's a brief, impactful event that sets the block in motion. The resulting velocity (v) immediately after the push depends on the magnitude of the impulse and the mass of the block (v = J/m).

Motion After the Push: Friction Takes Center Stage

Once the push ends, the block is no longer under the influence of Fpush. However, it now possesses an initial velocity (v) and is moving across the surface. This is where friction becomes the dominant force.

• Friction: Friction is a force that opposes motion between two surfaces in contact. It arises from microscopic irregularities and interactions between the surfaces.

• Types of Friction: There are two main types:

  • Static Friction: This force prevents an object from starting to move. It's what keeps the block at rest before the push. Static friction has a maximum value it can exert before being overcome.
  • Kinetic Friction: This force opposes the motion of an object already in motion. It's the force that slows down the block after the push. Kinetic friction is generally less than static friction.

• Coefficient of Friction (μ): The magnitude of the friction force is proportional to the normal force (N) and is determined by the coefficient of friction (μ), which is a dimensionless number that depends on the materials of the two surfaces in contact.

  • Kinetic Friction Force (Fk): Fk = μk * N, where μk is the coefficient of kinetic friction.

After the push, the net force acting on the block is primarily the kinetic friction force, acting in the opposite direction of the block's motion.

Deceleration and Stopping Distance

Since the net force is now the friction force opposing the motion, the block experiences deceleration (negative acceleration).

• Newton's Second Law (Again): Applying Newton's Second Law, we have -Fk = ma, where 'a' is now the deceleration.
• Calculating Deceleration: a = -Fk / m = - (μk * N) / m. Since N = mg (where g is the acceleration due to gravity), we can simplify this to a = -μk * g.

The deceleration is constant (assuming the coefficient of kinetic friction remains constant). Using kinematics equations, we can determine the stopping distance (d) – the distance the block travels before coming to a complete stop.

• Kinematics Equation: v^2 = u^2 + 2as, where:

  • v = final velocity (0 m/s in this case, as the block stops)
  • u = initial velocity (the velocity immediately after the push)
  • a = acceleration (which is the negative deceleration due to friction, -μk * g)
  • s = displacement (which is the stopping distance, d)

• Solving for Stopping Distance: 0 = u^2 + 2(-μk * g)d. Rearranging, we get d = u^2 / (2 * μk * g).

This equation shows that the stopping distance is proportional to the square of the initial velocity and inversely proportional to the coefficient of kinetic friction and the acceleration due to gravity.

Energy Considerations: Work and Energy Dissipation

Another way to analyze this scenario is through the lens of energy.

• Kinetic Energy: After the push, the block possesses kinetic energy (KE), given by KE = 1/2 * mv^2, where v is the initial velocity.
• Work-Energy Theorem: The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In this case, the work done by friction is negative, as it opposes the motion and reduces the kinetic energy.
• Work Done by Friction: The work done by friction (Wf) is given by Wf = -Fk * d = -μk * N * d = -μk * mg * d, where d is the stopping distance.

As the block slides to a stop, its kinetic energy is gradually converted into thermal energy due to the friction between the block and the surface. This thermal energy is dissipated into the surroundings, slightly increasing the temperature of both the block and the surface.

Factors Affecting the Motion

Several factors can influence the motion of the block:

• Magnitude of the Push: A stronger push imparts a greater impulse, resulting in a higher initial velocity and a longer stopping distance.
• Mass of the Block: A heavier block requires a greater impulse to achieve the same initial velocity. It also experiences a larger friction force (since the normal force is greater), but the deceleration is the same (a = -μk * g).
• Coefficient of Friction: A higher coefficient of friction results in a greater friction force and a shorter stopping distance.
• Surface Properties: The roughness and materials of the block and the surface significantly affect the coefficient of friction. A smoother surface will generally have a lower coefficient of friction.
• Angle of the Surface: If the surface is inclined, the analysis becomes more complex as the gravitational force needs to be resolved into components parallel and perpendicular to the surface. This affects both the normal force and the net force acting on the block.
• Air Resistance: At higher velocities, air resistance can become a significant factor, adding another force opposing the motion. However, for typical block-sliding scenarios at relatively low speeds, air resistance is often negligible.

A More Complex Scenario: Push with Sustained Force

Let's consider a variation where the push is not instantaneous but a sustained force applied over a longer period. In this case, the analysis changes slightly.

• Constant Acceleration: If the force Fpush is constant and applied for a duration Δt, the block experiences constant acceleration during that time.
• Net Force: The net force during the push is now Fpush - Fk (assuming the block is already in motion and experiencing kinetic friction).
• Acceleration: The acceleration is given by a = (Fpush - Fk) / m = (Fpush - μk * mg) / m.
• Velocity as a Function of Time: The velocity of the block increases linearly with time during the push: v(t) = u + at, where u is the initial velocity (which could be zero if the block starts from rest) and t is the time elapsed since the push began.

After the sustained push ends, the analysis reverts to the previous scenario where the only force acting on the block is friction, causing it to decelerate and eventually stop. The main difference is that the initial velocity after the push will be higher due to the continuous application of force.

Real-World Applications and Examples

The principles governing the motion of a block given a quick push are applicable to numerous real-world situations:

• Sports: Understanding friction and momentum is crucial in sports like hockey (the puck sliding on ice), bowling (the ball rolling down the lane), and baseball (the sliding of a player into a base).
• Automotive Engineering: The braking system in a car relies on friction to decelerate the vehicle. Engineers carefully design braking systems to maximize friction while preventing skidding.
• Manufacturing: Conveyor belts utilize friction to move objects along a production line. The coefficient of friction between the belt and the objects must be sufficient to prevent slippage.
• Everyday Life: Even simple actions like pushing a box across the floor or sliding a book across a table involve these fundamental physics principles.

Numerical Examples

Let's illustrate these concepts with a few numerical examples:

Example 1:

A 2 kg block is given a quick push on a horizontal surface with a coefficient of kinetic friction of 0.3. The push imparts an initial velocity of 5 m/s. Calculate the stopping distance.

• Given: m = 2 kg, μk = 0.3, u = 5 m/s, g = 9.8 m/s²
• Stopping Distance: d = u^2 / (2 * μk * g) = (5 m/s)^2 / (2 * 0.3 * 9.8 m/s²) ≈ 4.25 meters

Example 2:

A 5 kg block is at rest. A force of 10 N is applied horizontally for 2 seconds. The coefficient of kinetic friction is 0.2. Calculate the velocity of the block after 2 seconds.

• Given: m = 5 kg, Fpush = 10 N, Δt = 2 s, μk = 0.2, g = 9.8 m/s²
• Friction Force: Fk = μk * mg = 0.2 * 5 kg * 9.8 m/s² = 9.8 N
• Net Force: Fnet = Fpush - Fk = 10 N - 9.8 N = 0.2 N
• Acceleration: a = Fnet / m = 0.2 N / 5 kg = 0.04 m/s²
• Final Velocity: v = u + at = 0 + (0.04 m/s²) * 2 s = 0.08 m/s

Example 3:

A 1 kg block slides down an inclined plane at 30 degrees. The coefficient of kinetic friction is 0.1. Calculate the acceleration of the block.

• Given: m = 1 kg, θ = 30°, μk = 0.1, g = 9.8 m/s²
• Component of Gravity Parallel to the Plane: Fg_parallel = mg * sin(θ) = 1 kg * 9.8 m/s² * sin(30°) = 4.9 N
• Component of Gravity Perpendicular to the Plane: Fg_perpendicular = mg * cos(θ) = 1 kg * 9.8 m/s² * cos(30°) ≈ 8.49 N
• Normal Force: N = Fg_perpendicular ≈ 8.49 N
• Friction Force: Fk = μk * N = 0.1 * 8.49 N ≈ 0.849 N
• Net Force: Fnet = Fg_parallel - Fk = 4.9 N - 0.849 N ≈ 4.051 N
• Acceleration: a = Fnet / m = 4.051 N / 1 kg ≈ 4.051 m/s²

Common Misconceptions

• Friction Always Stops Motion Immediately: Friction slows down motion, but it doesn't necessarily stop it instantly. The block will continue to move until all its kinetic energy is dissipated by friction.
• A Constant Push Always Results in Constant Velocity: A constant push only results in constant velocity if the net force is zero, which means the push force must be equal to the friction force. If the push force is greater than the friction force, the block will accelerate.
• Ignoring Static Friction: Static friction plays a crucial role in preventing the block from moving initially. It must be overcome by the applied force for the block to start moving.
• Assuming Friction is Constant: The coefficient of kinetic friction can vary depending on factors like the speed of the block, the temperature of the surfaces, and the presence of lubricants. In many simplified analyses, we assume a constant coefficient of friction for ease of calculation.

Conclusion

The seemingly simple act of giving a block a quick push unveils a rich tapestry of physics principles. Understanding the interplay of forces, inertia, friction, and energy allows us to predict and explain the subsequent motion of the block. From calculating stopping distances to analyzing energy dissipation, this scenario provides a valuable foundation for understanding more complex physical phenomena in various fields of science and engineering. By carefully considering the factors involved and applying the appropriate equations, we can gain a deeper appreciation for the elegance and power of classical mechanics.