/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 A \(0.30 \mathrm{~kg}\) ladle sl... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 48

A \(0.30 \mathrm{~kg}\) ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring \((k=500 \mathrm{~N} / \mathrm{m})\) whose other end is fixed. The ladle has a kinetic energy of \(10 \mathrm{~J}\) as it passes through its equilibrium position (the point at which the spring force is zero). (a) At what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) At what rate is the spring doing work on the ladle when the spring is compressed \(0.10 \mathrm{~m}\) and the ladle is moving away from the equilibrium position?

Short Answer

Expert verified
(a) The work rate is 0 W at equilibrium. (b) The work rate is -408 W when the spring is compressed 0.10 m.

Step by step solution

01

Analyze Given Information

We have a spring with spring constant \( k = 500 \,\mathrm{N/m} \) and a ladle with mass \( m = 0.30 \,\mathrm{kg} \). When the ladle is at the equilibrium position, its kinetic energy \( KE = 10 \,\mathrm{J} \). We need to find the work rate (power) at equilibrium and when the spring is compressed by \( 0.10 \,\mathrm{m} \).
02

Understand Work Rate (Power) at Equilibrium

The work rate is the derivative of work with respect to time: \( P = \dfrac{dW}{dt} \). At the equilibrium position, the force and displacement in the spring are zero, but kinetic energy provides velocity for calculating power.
03

Calculate Velocity at Equilibrium

Use the kinetic energy formula \( KE = \frac{1}{2}mv^2 \). Solving for velocity, \( v = \sqrt{\frac{2 \cdot KE}{m}} \). Substitute \( KE = 10 \mathrm{~J} \) and \( m = 0.30 \mathrm{~kg} \) to find \( v = \sqrt{\frac{2 \times 10}{0.30}} \approx 8.16 \,\mathrm{m/s} \).
04

Calculate Force by Spring at Equilibrium

At equilibrium, the spring does not exert any force since the spring force is defined as \( F = -kx \) and \( x = 0 \) here, so \( F = 0 \). Hence, the work rate \( P = Fv = 0 \times 8.16 = 0 \,\mathrm{W} \).
05

Calculate Power when Spring is Compressed by 0.10 m

First, find the force exerted by the spring: \( F = -kx = -500 \times 0.10 = -50 \,\mathrm{N} \). Power \( P = Fv \), assuming the speed remains \( 8.16 \,\mathrm{m/s} \). Hence, \( P = -50 \times 8.16 = -408 \,\mathrm{W} \). The negative sign indicates the spring is doing work on the ladle as it is moving away from the equilibrium.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is an important concept in spring dynamics because it helps us understand how energy is transferred between a moving object and the spring it interacts with.
The kinetic energy of a moving object can be calculated using the formula: \[KE = \frac{1}{2} mv^2\]where:
  • \( KE \) is the kinetic energy,
  • \( m \) is the mass of the object,
  • \( v \) is the velocity of the object.
In this exercise, the ladle has a kinetic energy of 10 J at the equilibrium position, indicating how fast it is moving as it passes through this point. Understanding how kinetic energy functions at different stages of motion helps in analyzing systems like springs, where motion continually cycles between potential and kinetic energy.
Spring Constant
The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It is crucial in calculating the force exerted by a spring when it is stretched or compressed.
The relationship between the spring force \( F \) and displacement from its equilibrium position \( x \) is given by Hooke's Law:\[F = -kx\]
  • The negative sign indicates that the spring force is a restoring force, acting in the opposite direction to the displacement.
  • A higher spring constant means a stiffer spring, which requires more force to achieve the same displacement.
In this example, the spring constant is 500 N/m. This indicates that for every meter the spring is compressed or stretched from its equilibrium position, a force of 500 Newtons is exerted by the spring.
Work Rate
The work rate, or power, defines how quickly work is being done by a force over time. In physics, power is the rate at which energy is transferred or converted. For springs, this involves transferring potential energy stored in the spring into kinetic energy of the moving object, or vice versa.
Power \( P \) is mathematically expressed as:\[P = \dfrac{dW}{dt} = Fv\]where:
  • \( P \) is the power,
  • \( dW \) is the differential work done,
  • \( dt \) is the change in time,
  • \( F \) is the force applied,
  • \( v \) is the velocity of the object.
In this exercise, when the ladle is at the equilibrium position, the spring does not exert force, so the power is zero. When compressed, the spring exerts a force, and the power is calculated to be -408 W, indicating work is being done against the spring's direction.
Equilibrium Position
The equilibrium position of a spring system is the point where the net force is zero, and for ideal springs, it is the position where the spring is neither compressed nor stretched.
In the context of the exercise, as the ladle passes through this position, it means:
  • The spring force is zero at equilibrium, hence no work is done by the spring.
  • All kinetic energy is transferred into the system's motion due to the absence of spring force.
Understanding equilibrium is key in analyzing spring systems as it represents the point of balance, where potential energy is lowest, and the system displays maximum kinetic activity. As the motion proceeds beyond the equilibrium, potential energy increases as the spring compresses or stretches.

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Most popular questions from this chapter

An explosion at ground level leaves a crater with a diameter that is proportional to the energy of the explosion raised to the \(\frac{1}{3}\) power; an explosion of 1 megaton of TNT leaves a crater with a \(1 \mathrm{~km}\) diameter. Below Lake Huron in Michigan there appears to be an ancient impact crater with a \(50 \mathrm{~km}\) diameter. What was the kinetic energy associated with that impact, in terms of (a) megatons of TNT (1 megaton yields \(\left.4.2 \times 10^{15} \mathrm{~J}\right)\) and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)? (Ancient meteorite or comet impacts may have significantly altered Earth's climate and contributed to the extinction of the dinosaurs and other life-forms.)

(a) In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN}\), was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm}\), how much work did her force do on the car?

A helicopter lifts a \(72 \mathrm{~kg}\) astronaut \(15 \mathrm{~m}\) vertically from the ocean by means of a cable. The acceleration of the astronaut is \(g / 10 .\) How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

A force \(\vec{F}=\left(c x-3,00 x^{2}\right) \hat{\mathrm{i}}\) acts on a particle as the particle moves along an \(x\) axis, with \(\vec{F}\) in newtons, \(x\) in meters, and \(c\) a constant. At \(x=0\), the particle's kinetic energy is \(20.0 \mathrm{~J} ;\) at \(x=3.00\) \(\mathrm{m}\), it is \(11.0 \mathrm{~J}\). Find \(c\).

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(900 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(8.0\) \(\mathrm{m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0\) \(\mathrm{m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{c}) 2.0 \mathrm{~m} / \mathrm{s} ?\)
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