Problem 57 - University Physics with Modern Physics

Question

A machine part of mass \(m\) is attached to a horizontal ideal spring of force constant \(k\) that is attached to the edge of a friction-free horizontal surface. The part is pushed against the spring compressing it a distance \(x_{0},\) and then released from rest. Find the maximum (a) speed and (b) acceleration of the machine part. (c) Where in the motion do the maxima in parts (a) and (b) occur? (d) What will be the maximum extension of the spring? (e) Describe the subsequent motion of this machine part. Will it ever stop permanently?

Step-by-Step Solution

Answer

(a) Maximum speed: \(v_{max} = \sqrt{\frac{k}{m}} x_0\). (b) Maximum acceleration: \(a_{max} = \frac{k}{m} x_0\). (c) Speed max at equilibrium; acceleration max at max extension/compression. (d) Max extension: \(x_0\). (e) Part oscillates indefinitely.

Step 1: Understand the Scenario

The problem involves a mass attached to a spring on a frictionless surface. When compressed and released, it undergoes simple harmonic motion. Our task is to find the maximum speed, acceleration, and other aspects of the motion.

Step 2: Identify Maximum Speed

The maximum speed occurs when the potential energy (stored in the spring) is completely converted into kinetic energy. At this point, the mass passes through the equilibrium position. Using conservation of energy: \[\frac{1}{2} k x_0^2 = \frac{1}{2} m v_{max}^2\]Solving for \(v_{max}\):\[v_{max} = \sqrt{\frac{k}{m}} x_0\]

Step 3: Determine Maximum Acceleration

The maximum acceleration occurs when the spring force is at its maximum, which is at maximum compression or extension (meaning at \(x_0\)). The maximum acceleration is given by Hooke's Law: \[F = ma = kx_0\]Thus, \[a_{max} = \frac{k}{m} x_0\]

Step 4: Maxima Occurrence

(a) Maximum speed occurs at the equilibrium position where potential energy is converted to kinetic energy. (b) Maximum acceleration occurs at maximum compression or extension \( x = \pm x_0 \).

Step 5: Determine Maximum Extension

Due to conservation of energy, the maximum extension of the spring is equal to the maximum compression, which is \(x_0\).

Step 6: Describe Subsequent Motion

The motion will be simple harmonic. The part oscillates back and forth between the maximum compression and maximum extension without ever stopping permanently due to the absence of friction.

Key Concepts

Energy Conservation

In simple harmonic motion, energy conservation plays a critical role. The total mechanical energy of the system is constant. This means that the sum of kinetic energy and potential energy does not change over time.

When the spring is fully compressed with the mass at rest, all the energy is in the form of potential energy, given by

• Potential Energy (PE) = \( \frac{1}{2} k x_0^2 \)

As the spring is released, this potential energy is converted into kinetic energy (KE) as the mass moves towards the equilibrium position. Maximal speed occurs here when all energy is kinetic:

• Kinetic Energy (KE) = \( \frac{1}{2} m v_{max}^2 \)

Thus, conservation of energy can be stated as:

• \( \frac{1}{2} k x_0^2 = \frac{1}{2} m v_{max}^2 \)
• Simplifying gives us \( v_{max} = \sqrt{\frac{k}{m}} x_0 \)

Hooke's Law

Hooke's Law provides the force exerted by a spring when it is compressed or extended. This law is crucial in calculating forces in simple harmonic motion. Hooke's Law states that the force \( F \) exerted by the spring is directly proportional to the displacement \( x \) from the equilibrium position:\[ F = -kx \] - The negative sign indicates that the force is a restoring force, acting in the opposite direction of displacement.- Maximum acceleration in this motion occurs where the spring's force is greatest. When the mass is at maximum compression or extension, the magnitude of force is:

• \( F = kx_0 \)
• Acceleration \( a = \frac{F}{m} = \frac{k}{m} x_0 \)

Maximum Speed

In the context of an oscillating system, maximum speed is obtained as the object passes through the equilibrium position. This is where potential energy has fully converted to kinetic energy. The speed is highest because:- The spring is neither compressed nor extended, meaning no force acts to slow down the mass at that point.Using energy conservation, we find that:

• Maximum speed: \( v_{max} = \sqrt{\frac{k}{m}} x_0 \)

This value shows how the force constant \( k \) and mass \( m \) influence the speed.

Maximum Acceleration

The maximum acceleration occurs at the points of maximum spring compression or extension. At these points, the spring's restoring force is at its peak due to Hooke's Law:

• The largest force \( F = kx_0 \) determines the maximum acceleration: \( a_{max} = \frac{k}{m} x_0 \)

This acceleration can be understood as follows:- It acts to restore the system to its equilibrium.- It is highest where displacement is the greatest from equilibrium, resulting in maximum force.Understanding maximum acceleration is essential, especially in engineering scenarios where minimizing unwanted oscillations is critical.