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Chapter 7: Problem 57

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?

Short Answer

Expert verified
(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 by step solution

01

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.
02

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 \(v_{max}\):\[v_{max} = \sqrt{\frac{k}{m}} x_0\]
03

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, \[a_{max} = \frac{k}{m} x_0\]
04

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 \( x = \pm x_0 \).
05

Determine Maximum Extension

Due to conservation of energy, the maximum extension of the spring is equal to the maximum compression, which is \(x_0\).
06

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.

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Key Concepts

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

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.

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Most popular questions from this chapter

Bungee Jump. A bungee cord is 30.0 \(\mathrm{m}\) long and, when stretched a distance \(x,\) it exerts a restoring force of magnitude \(k x\) Your father-in- law (mass 95.0 \(\mathrm{kg}\) ) stands on a platform 45.0 \(\mathrm{m}\) above the ground, and one end of the cord is tied securely to his ankle and the other end to the platform. You have promised him that when he steps off the platform he will fall a maximum distance of only 41.0 \(\mathrm{m}\) before the cord stops him. You had several bungee cords to select from, and you tested them by stretching them out, tying one end to a tree, and pulling on the other end with a force of 380.0 \(\mathrm{N}\) . When you do this, what distance will the bungee cord that you should select have stretched?

You are designing a delivery ramp for crates containing exercise equipment. The \(1470-\mathrm{N}\) crates will move at 1.8 \(\mathrm{m} / \mathrm{s}\) at the top of a ramp that slopes downward at \(22.0^{\circ} .\) The ramp exerts a \(550-\mathrm{N}\) kinetic friction force on each crate, and the maximum static friction force also has this value. Each crate will compress a spring at the bottom of the ramp and will come to rest after traveling a total distance of 8.0 \(\mathrm{m}\) along the ramp. Once stopped, a crate must not rebound back up the ramp. Calculate the force constant of the spring that will be needed in order to meet the design criteria.

A 10.0 -kg microwave oven is pushed 8.00 \(\mathrm{m}\) up the sloping surface of a loading ramp inclined at an angle of \(36.9^{\circ}\) above the horizontal, by a constant force \(\vec{F}\) with a magnitude 110 \(\mathrm{N}\) and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250 . (a) What is the work done on the oven by the force \(\vec{F} ?\) (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), \((b),\) and (c) to calculate the increase in the oven's kinetic energy. \((e)\) Use \(\Sigma \overrightarrow{\boldsymbol{F}}=m \overrightarrow{\boldsymbol{a}}\) to calculate the acceleration of the oven. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after traveling 8.00 \(\mathrm{m}\) . From this, compute the increase in the oven's kinetic energy, and compare it to the answer you got in part (d).

A variable force \(\overrightarrow{\boldsymbol{k}}\) is maintained tangent to a frictionless semicircular surface (Fig. 7.41\() .\) By slow variations in the force, a block with weight \(w\) is moved, and the spring to which it is attached is stretched from position 1 to position \(2 .\) The spring has negligible mass and force constant \(k .\) The end of the spring moves in an are of radius \(a\) . Calculate the work done by the force \(\vec{F}\) .

A 0.60\(\cdot \mathrm{kg}\) book slides on a horizontal table. The kinetic friction force on the book has magnitude 1.2 \(\mathrm{N}\) . (a) How much work is done on the book by friction during a displacement of 3.0 \(\mathrm{m}\) to the left? (b) The book now slides 3.0 \(\mathrm{m}\) to the right, returning to its starting point. During this second 3.0 \(\mathrm{m}\) displacement, how much work is done on the book by friction? (c) What is the total work done on the book by friction during the complete round trip? (d) On the basis of your answer to part (c), would you say that the friction force is conservative or nonconservative? Explain.
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