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Chapter 7: Problem 68

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 \(\vec{F}\) .

Short Answer

Expert verified
Work done by the force is \(\frac{1}{2} k a^2 \pi^2\).

Step by step solution

01

Understanding the System

We are given a block attached to a spring being moved by a force along a semicircular path with radius \(a\). The spring is stretched from position 1 to position 2 as the block moves. The force \(\vec{F}\) applied is always tangent to the path and varies slowly.
02

Calculating Work Done by Force with Springs

The work done by the force \(\vec{F}\) is equal to the change in mechanical energy of the system. For a spring, the change in potential energy \(U\) when it is stretched from an initial position \(x_1\) to a final position \(x_2\) is given by \( U = \frac{1}{2}k x_2^2 - \frac{1}{2}k x_1^2 \), where \(k\) is the spring constant.
03

Assessing Path of Spring

As the spring is stretched along the semicircular path, its displacement corresponds to the arc length. The arc length \(s\) for a semicircular path with radius \(a\) and subtended angle \(\theta\) in radians is \(s = a\theta\). For a half-circle, \(\theta = \pi\), and the arc length is \(s = a\pi\).
04

Relating Arc Length to Work Done

Assuming the spring displaces linearly along the path, the change in its potential energy is related to the arc length. The work done by the force will convert into spring potential energy, which can be considered as a linear function of distance halted via arc length \(a\pi\).
05

Calculating Work Done

Using the energy relation, the work done by the tangential force \(W = \Delta U\) is given by \(W = \frac{1}{2} k a^2 \pi^2\) as it changes potential energy from position 1 to position 2. Notice there are no heights considered since the movement is frictionless and contained within a semicircular path.

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Key Concepts

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

Spring Force
When you hear "spring force," think of it like a rubber band or a spring that wants to return to its original size after being stretched or compressed. In a spring, the force that resists these changes is described by Hooke's Law:
  • The force exerted by the spring, described as \( F = -kx \)
  • Where
    • \( F \) is the force from the spring,
    • \( k \) is the spring constant (a measure of stiffness),
    • and \( x \) is the displacement from the spring's resting position.
The negative sign in the equation represents that the spring force acts in the opposite direction of the displacement. This spring force is constantly acting to bring the block back to its starting position.
Variable Force
A variable force is one that changes in magnitude and/or direction over time. Consider gently pushing a swinging pendulum: your force changes direction as the pendulum swings back and forth. In our scenario,
  • the force \( \vec{F} \) is always tangent to the semicircular path and
  • changes slowly to ensure it aligns with the path of motion.
This slow variation is necessary for controlling the motion precisely, especially because the force adjusting the spring also varies with displacement. Recognizing a variable force requires understanding that the force isn鈥檛 constant but adjusts its application based on the position or movement of the system it affects.
Semicircular Motion
Semicircular motion refers to movement along a half-circle or arc. In this problem, the block attached to a spring follows this type of path鈥攖hink of it like a swing on a playground that traces out a semi-circle.
  • The radius \( a \) describes how "wide" the semicircular path is.
  • The arc length \( s = a\theta \) quantifies how far around the circle the object moves, where in a full semicircle, \( \theta \) is \( \pi \) (180 degrees).
As the block moves, the spring stretches following this arc path, which inherently affects the work needed to move the block, because the path affects how much the spring stretches.
Potential Energy Change
The change in potential energy
  • reflects how energy is stored in the spring as it stretches and relocates along its path.
  • When considering the work done to move the block鈥攁 process transforming kinetic energy into potential energy鈥攚e focus on the initial and final states of energy.
For the spring in question, as the block moves from one point to another on its path, it accumulates potential energy described by: \[U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2\]where \( x_1 \) and \( x_2 \) are initial and final positions of the spring. This change accounts for work done by the force \( \vec{F} \) and equates the transformation of energy from one form to another within the system, specifically into spring potential energy due to displacement.

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Most popular questions from this chapter

In one day, a \(75-\mathrm{kg}\) mountain climber ascends from the 1500 -m level on a vertical cliff to the top at 2400 \(\mathrm{m}\) . The next day, she descends from the top to the base of the cliff, which is at an elevation of 1350 \(\mathrm{m}\) . What is her change in gravitational potential energy (a) on the first day and \((b)\) on the second day?

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Legal Physics. In an auto accident, a car hit a pedestrian and the driver then slammed on the brakes to stop the car. During the subsequent trial, the driver's lawyer claimed that he was obeying the posted 35 \(\mathrm{mi}\) h speed limit, but that the legal speed was too high to allow him to see and react to the pedestrian in time. You have been called in as the state's expert witness. Your investigation of the accident found that the skid marks made while the brakes were applied were 280 \(\mathrm{ft}\) long, and the tread on the tires produced a coefficient of kinetic friction of 0.30 with the road. (a) In your testimony in court, will you say that the driver was obeying the posted speed? You must be able to back up your conclusion with clear reasoning because one of the lawyers will surely cross-examine you. (b) If the driver's speeding ticket were \(\$ 10\) for each mile per hour be was driving above the posted speed limit, would he have to pay a fine? If so, how much would it be?

A force parallel to the \(x\) -axis acts on a particle moving along the \(x\) -axis. This force produces potential energy \(U(x)\) given by \(U(x)=\alpha x^{4},\) where \(\alpha=1.20 \mathrm{J} / \mathrm{m}^{4} .\) What is the force (magnitude and direction) when the particle is at \(x=-0.800 \mathrm{m} ?\)
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