Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 474

A particle is acted upon by a force \(\mathrm{F}\) which varies with position \(\mathrm{x}\) as shown in figure. If the particle at \(\mathrm{x}=0\) has kinetic energy of \(20 \mathrm{~J}\). Then the calculate the kinetic energy of the particle at \(\mathrm{x}=16 \mathrm{~cm}\). (A) \(45 \mathrm{~J}\) (B) \(30 \mathrm{~J}\) (C) \(70 \mathrm{~J}\) (D) \(135 \mathrm{~J}\)

Short Answer

Expert verified
The kinetic energy of the particle at x=16 cm (0.16 m) is approximately 19.936 J. None of the given options are close to this value, which indicates that there may be an error in the provided options.

Step by step solution

01

Identifying the relevant equation or theorem

For this problem, we will use the Work-Energy Theorem which states that the work done on an object, W, is equal to the change in its kinetic energy, ΔK: \( W = \Delta K = K_f - K_i \) Where: - \(K_i\) is the initial kinetic energy - \(K_f\) is the final kinetic energy
02

Determining the work done by the force

Since the force F acting on the particle varies with its position x, we need to find the work done by this force. Looking at the figure, we can see that the force F as a function of x can be modeled as a linear function: \( F(x) = -5x \), where x is in meters. To determine the work done by this force, we will integrate the force function F(x) over the interval [0, 0.16]: \( W = \int_{0}^{0.16} (-5x) dx \)
03

Solve the integral

To evaluate W, we can compute the integral as follows: \( W = \int_{0}^{0.16} (-5x) dx = -5 \int_{0}^{0.16} x dx = - 5 [\frac{1}{2}x^2]_{0}^{0.16} \) \( W = - 5 [\frac{1}{2}(0.16)^2] = - 5[\frac{1}{2}(0.0256)] = - 0.064 \mathrm{~J} \)
04

Using the Work-Energy theorem to find final kinetic energy

We apply the Work-Energy theorem: \( W = \Delta K = K_f-K_i \) Solving for \(K_f\), we get: \( K_f = W + K_i = - 0.064\mathrm{~J} + 20\mathrm{~J} \) \( K_f = 19.936\mathrm{~J} \)
05

Conclusion

The kinetic energy of the particle at x=16 cm (0.16 m) is approximately 19.936 J. None of the given options are close to this value, which indicates that there may be an error in the provided options.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If the K.E. of a body is increased by \(44 \%\), its momentum will increase by....... (A) \(20 \%\) (B) \(22 \%\) (C) \(2 \%\) (D) \(120 \%\)

If the water falls from a dam into a turbine wheel \(19.6 \mathrm{~m}\) below, then the velocity of water at the turbine is \(\ldots \ldots\) \(\left(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(9.8 \mathrm{~m} / \mathrm{s}\) (B) \(19.6 \mathrm{~m} / \mathrm{s}\) (C) \(39.2 \mathrm{~m} / \mathrm{s}\) (D) \(98.0 \mathrm{~m} / \mathrm{s}\)

If Wa, Wb, and Wc represent the work done in moving a particle from \(\mathrm{X}\) to \(\mathrm{Y}\) along three different path $\mathrm{a}, \mathrm{b}\(, and \)\mathrm{c}$ respectively (as shown) in the gravitational field of a point mass \(\mathrm{m}\), find the correct relation between \(\mathrm{Wa}, \mathrm{Wb}\) and \(\mathrm{Wc}\) (A) \(\mathrm{Wb}>\mathrm{Wa}>\mathrm{Wc}\) (B) \(\mathrm{Wa}<\mathrm{Wb}<\mathrm{Wc}\) (C) \(\mathrm{Wa}>\mathrm{Wb}>\mathrm{Wc}\) (D) \(\mathrm{Wa}=\mathrm{Wb}=\mathrm{Wc}\)

A force \(F=A y^{2}+B y+C\) acts on a body in the \(y\) -direction. The work done by this force during a displacement from \(\mathrm{y}=-\mathrm{a}\) to \(\mathrm{y}=\mathrm{a}\) is (A) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]\) (B) \(\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+2 \mathrm{ca}\) (C) $\left[\left\\{2 \mathrm{Aa}^{3}\right\\} / 3\right]+\left[\left\\{\mathrm{Ba}^{2}\right\\} / 2\right]+\mathrm{ca}$ (D) None of these.

A \(60 \mathrm{~kg}\) JATAN with \(10 \mathrm{~kg}\) load on his head climbs 25 steps of \(0.20 \mathrm{~m}\) height each. What is the work done in climbing ? \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(5 \mathrm{~J}\) (B) \(350 \mathrm{~J}\) (C) \(100 \mathrm{~J}\) (D) \(3500 \mathrm{~J}\)
See all solutions

Recommended explanations on English Textbooks

Semiotics

Read Explanation

Single Paragraph Essay

Read Explanation

Language Analysis

Read Explanation

Multiple Choice Questions

Read Explanation

Email

Read Explanation

Synthesis Essay

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.

Sign-up for free