/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

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 rate is 0 W at equilibrium. (b) The rate is -365 W when compressed 0.10 m.

Step by step solution

01

Understand the Spring Force and Work Rate

The rate at which the spring does work is synonymous with the power exerted by the spring. At the equilibrium position, the velocity of the ladle is at its maximum, and thus the power exerted by the spring is zero because the spring force is zero. This is because the spring is neither compressed nor stretched, so the work being done by the spring on the ladle is zero at the equilibrium point.
02

Calculate the Work Rate at Equilibrium Position

Since the spring is neither compressed nor stretched at the equilibrium position, the spring force and thus the work done by the spring on the ladle is zero. Therefore, the power, or rate of work, exerted by the spring at this point is zero.
03

Use Hooke's Law for Spring Force Calculation

As the spring is compressed 0.10 meters, the spring force can be calculated using Hooke's Law: \[ F = -kx \]where \( F \) is the spring force, \( k = 500 \, \text{N/m} \) is the spring constant, and \( x = 0.10 \, \text{m} \) is the compression distance.Thus:\[ F = -500 \, \text{N/m} \times 0.10 \, \text{m} = -50 \, \text{N} \]
04

Power Calculation for Compressed Spring

To find the rate at which work is done (power) when the spring is compressed, we need the velocity of the ladle. Given the initial kinetic energy (\( KE = 10 \, \text{J} \)) at equilibrium and the potential energy stored in the spring when compressed (\( PE = \frac{1}{2}kx^2 \)), potential energy is:\[ PE = \frac{1}{2} \times 500 \, \text{N/m} \times (0.10 \, \text{m})^2 = 2.5 \, \text{J} \]Using the conservation of mechanical energy:\[ KE_{final} = KE_{initial} - PE \]\[ KE_{final} = 10 \, \text{J} - 2.5 \, \text{J} = 7.5 \, \text{J} \]Find velocity \( v \) using \( KE = \frac{1}{2}mv^2 \):\[ 7.5 \, \text{J} = \frac{1}{2} \times 0.30 \, \text{kg} \times v^2 \]\[ v = \sqrt{ \frac{15}{0.30} } \approx 7.3 \, \text{m/s} \]Power \( P \) is given by the product of force and velocity:\[ P = F \times v = -50 \, \text{N} \times 7.3 \, \text{m/s} = -365 \, \text{W} \]. The negative sign indicates that the spring is doing work on the ladle.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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. In the context of our ladle and spring system, the kinetic energy of the ladle is at its maximum when it passes through the equilibrium position. This occurs because the spring force, which could otherwise act to oppose the motion of the ladle, is zero at equilibrium.
As given in the problem, the kinetic energy of the ladle is 10 J at the equilibrium position. This energy gives us the speed of the ladle at this point. The equation for kinetic energy is \[ KE = \frac{1}{2}mv^2 \]where:
  • \( KE \) is the kinetic energy (10 J in this scenario),
  • \( m \) is the mass of the ladle (0.30 kg), and
  • \( v \) is the velocity.
From this formula, one can solve for the velocity \( v \) to understand how fast the ladle is moving when it is not affected by the spring force.
Spring Force
Spring force, also known as restoring force, is the force exerted by a compressed or stretched spring upon any object that is attached to it. According to Hooke's Law, this force is proportional to the displacement of the spring from its natural length and can be calculated by the formula\[ F = -kx \]Here:
  • \( F \) is the spring force,
  • \( k \) is the spring constant, which measures the stiffness of the spring (500 N/m), and
  • \( x \) is the displacement from the equilibrium position (in this case, 0.10 m).
When the spring is compressed, it tries to return to its equilibrium length by exerting a force in the opposite direction of the displacement. For instance, compressing the spring by 0.10 meters results in a force of 50 N directed forwards, indicating the spring is ready to push back.
Mechanical Energy Conservation
The principle of mechanical energy conservation states that in the absence of non-conservative forces (like friction), the total mechanical energy of a system remains constant. This total energy is the sum of kinetic energy and potential energy stored in the spring.
In this system, when the ladle is at the equilibrium position, all energy is kinetic because potential energy due to spring compression or expansion is zero. As the spring compresses (or stretches), potential energy builds up and kinetic energy decreases, or vice versa.
You can apply the conservation principle with: \[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]In our case, it simplifies the calculations as the potential energy at equilibrium is zero. By knowing that point values, like the initial kinetic energy of 10 J, you derive the energy distribution as the spring compresses 0.10 meters, computing changes in velocity and force dynamically as the ladle takes on different forms of energy along its path in the system.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A horse pulls a cart with a force of \(40 \mathrm{lb}\) at an angle of \(30^{\circ}\) above the horizontal and moves along at a speed of \(6.0 \mathrm{mi} / \mathrm{h} .\) (a) How much work does the force do in \(10 \mathrm{~min} ?\) (b) What is the average power (in horsepower) of the force?

A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) while force \(\vec{F}=(2 \mathrm{~N}) \hat{\mathrm{i}}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

A \(12.0 \mathrm{~N}\) force with a fixed orientation does work on a particle as the particle moves through the three-dimensional displacement \(\vec{d}=(2.00 \hat{\mathrm{i}}-4.00 \hat{\mathrm{j}}+3.00 \hat{\mathrm{k}}) \mathrm{m} .\) What is the angle between the force and the displacement if the change in the particle's kinetic energy is (a) \(+30.0 \mathrm{~J}\) and (b) \(-30.0 \mathrm{~J}\) ?

During spring semester at MIT, residents of the parallel buildings of the East Campus dorms battle one another with large catapults that are made with surgical hose mounted on a window frame. A balloon filled with dyed water is placed in a pouch attached to the hose, which is then stretched through the width of the room. Assume that the stretching of the hose obeys Hooke's law with a spring constant of \(100 \mathrm{~N} / \mathrm{m}\). If the hose is stretched by \(5.00 \mathrm{~m}\) and then released, how much work does the force from the hose do on the balloon in the pouch by the time the hose reaches its relaxed length?

If a car of mass \(1200 \mathrm{~kg}\) is moving along a highway at \(120 \mathrm{~km} / \mathrm{h}\), what is the car's kinetic energy as determined by someone standing alongside the highway?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.