/*! 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 118 A particle moves in a straight l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 118

A particle moves in a straight line with retardation proportion to its displacement. Its loss of kinetic energy for any displacement \(x\) is proportional to [2004] (A) \(x\) (B) \(e^{x}\) (C) \(x^{2}\) (D) \(\log _{e} x\)

Short Answer

Expert verified
The loss of kinetic energy for any displacement \(x\) is proportional to \(x^{2}\).

Step by step solution

01

Introduce the Variables

We are given that the retardation (negative acceleration) is proportional to the displacement (\(x\)). We can express this relationship as an equation: \(a = -kx\) Where, \(a\) is the acceleration, \(k\) is a constant of proportionality and \(x\) is the displacement.
02

Introduce Newton's Second Law

Newton's second law states that force acting on a body is equal to the mass times the acceleration of the body: \(F = ma\) Since the friction (retardation) is acting against the motion, we can replace the force \((F)\) with the negative of the force \((F_{r})\), i.e., \(-F_{r} = -ma\)
03

Relate the Retardation and Displacement

Using Step 1 and Step 2 results, we have: \(-F_{r} = -m(-kx)\) \(F_{r} = mkx\) Now let's consider the work-energy theorem to calculate the work done by this force (retardation) over the displacement.
04

Apply the Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy: \(W = \Delta K\) In our case, work (\(W\)) is done by the retarding force, and it is equal to the loss of kinetic energy. We can express the work done by the retarding force as: \(W = \int_{x_{1}}^{x_{2}} F_{r}dx\)
05

Calculate the Loss of Kinetic Energy as a Function of Displacement

Now we can substitute the expression for retarding force from step 3 and integrate to find the relationship between the loss of kinetic energy and displacement: \(\Delta K = \int_{x_{1}}^{x_{2}} mkxdx\) \(\Delta K = mk\int_{x_{1}}^{x_{2}} xdx\) \(\Delta K = mk\Big[\frac{1}{2}x^2\Big]_{x_{1}}^{x_{2}}\) Since we are ultimately interested in the loss of kinetic energy for "any" displacement \(x\), let's take \(x_{1} = 0\) and \(x_{2} = x\). Therefore: \(\Delta K = mk\Big[\frac{1}{2}x^2\Big]_{0}^{x}\) \(\Delta K = \frac{1}{2}mkx^{2}\) We can see that the loss of kinetic energy is proportional to \(x^{2}\). Hence, the correct answer is: (C) \(x^{2}\)

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 91影视!

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 wire suspended vertically from one of its ends is stretched by attaching a weight of \(200 \mathrm{~N}\) to the lower end. The weight stretches the wire by \(1 \mathrm{~mm}\). Then the elastic energy stored in the wire is \([\mathbf{2 0 0 3}]\) (A) \(0.2 \mathrm{~J}\) (B) \(10 \mathrm{~J}\) (C) \(20 \mathrm{~J}\) (D) \(0.1 \mathrm{~J}\)

A small block of mass \(m\) lying at rest at point \(P\) of a wedge having a smooth semi-circular track of radius \(R\). What should be the minimum value of horizontal acceleration \(a_{0}\) of wedge so that mass can just reach the point \(Q\) ? (A) \(g / 2\) (B) \(\sqrt{g}\) (C) \(g\) (D) Not possible

Two identical balls are projected, one vertically up and the other at an angle of \(30^{\circ}\) with the horizontal, with same initial speed. The potential energy at the highest point is in the ratio (A) \(4: 3\) (B) \(3: 4\) (C) \(4: 1\) (D) \(1: 4\)

This question has Statement 1 and Statement \(2 .\) Of the four choices given after the statements, choose the one that best describes the two statements. If two springs \(S_{1}\) and \(S_{2}\) of force constants \(k_{1}\) and \(k_{2}\), respectively, are stretched by the same force, it is found that more work is done on spring \(S_{1}\) than on spring \(S_{2}\) Statement 1: If stretched by the same amount, work done on \(S_{1}\) will be more than that on \(S_{2}\) Statement \(2: k_{1}

A proton is kept at rest. A positively charged particle is released from rest at a distance \(d\) in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a position. In the same time \(t\), the work done on the two moving charged particles is (A) Same as the force law is involved in the two experiments. (B) Less for the case of a positron, as the positron moves away more rapidly and the force on it weakens. (C) More in the case of positron, as the positron moves away a larger distance. (D) Same as the work done by charged particle on the stationary proton.
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Oscillations

Read Explanation

Kinematics Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Turning Points in Physics

Read Explanation

Electricity and Magnetism

Read Explanation
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.