Chapter 10: Q. 61 (page 259)

The potential energy for a particle that can move along the x-axis is $U = A x^{2} + B \sin (\frac{π x}{L})$ where A, B, and L are constants. What is the force on the particle at
(a) $X = 0,$
(b) $X = L / 2$ and
(c) $X = L$ ?

Short Answer

Expert verified

$(a) F_{x} = - B π / L (b) F_{x} = - AL (c) F_{x} = - 2 AL + \frac{Bπ}{L}$

Step by step solution

Given information (part a)

The potential energy of the particle along the x-axis is given as

$U = A x^{2} + B \sin (\frac{π x}{L})$

Explanation (part a)

The force F_X can be found by taking the negative derivative of potential energy with respect to position.

$U = A x^{2} + B \sin (\frac{π x}{L}) F_{x} = - \frac{d U}{d x} = - \frac{d}{d x} (A x^{2} + B \sin (\frac{π x}{L})) = - \frac{d}{d x} (A x^{2}) - \frac{d}{d x} (B \sin (\frac{π x}{L})) F_{x} = - 2 A x - B \cos (\frac{π x}{L}) \times (\frac{π}{L})$

At $X = 0$

$F_{x} = - 2 A \times 0 - B \cos (\frac{π (0)}{L}) \times (\frac{π}{L}) = 0 - B \cos (0) \times (\frac{π}{L}) = - B \times 1 \times (\frac{π}{L}) = - B (\frac{π}{L})$

Given information (part b)

The potential energy of the particle along the x-axis is given as

$U = A x^{2} + B \sin (\frac{π x}{L})$

Explanation (part b)

$A t X = L / 2$

localid="1649413461206" $F_{x} = - 2 A \times \frac{L}{2} - B \cos (\frac{π \times L}{2 L}) \times (\frac{π}{L}) = \frac{- 2 A L}{2} - B \cos (\frac{π}{2}) \times (\frac{π}{L}) = - A L - B (0) (\frac{π}{L}) = - A L$

Given information (part c)

The potential energy of the particle along the x-axis is given as

$U = A x^{2} + B \sin (\frac{π x}{L})$

Explanation (part c)

$A t X = L F_{x} = - 2 A \times L - B \cos ((\frac{π L}{L})) \times (\frac{π}{L}) = - 2 A L - B \cos (π) \times (\frac{π}{L}) = - 2 A L - B (- 1) (\frac{π}{L}) = - 2 A L + \frac{B π}{L}$

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!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

The potential energy for a particle that can move along the x-axis is $U = A x^{2} + B \sin (\frac{π x}{L})$ where A, B, and L are constants. What is the force on the particle at
(a) $X = 0,$
(b) $X = L / 2$ and
(c) $X = L$ ?

Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

Given information (part c)

Explanation (part c)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Measurements

Engineering Physics

Electricity and Magnetism

Solid State Physics

Rotational Dynamics

Force

Study anywhere. Anytime. Across all devices.

The potential energy for a particle that can move along the x-axis is U=Ax2+Bsin(πxL)where A, B, and L are constants. What is the force on the particle at(a) X=0,(b) X=L/2and(c) X=L?

Short Answer

Step by step solution

Given information (part a)

Explanation (part a)

Given information (part b)

Explanation (part b)

Given information (part c)

Explanation (part c)

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Measurements

Engineering Physics

Electricity and Magnetism

Solid State Physics

Rotational Dynamics

Force

Study anywhere. Anytime. Across all devices.

The potential energy for a particle that can move along the x-axis is $U = A x^{2} + B \sin (\frac{π x}{L})$ where A, B, and L are constants. What is the force on the particle at
(a) $X = 0,$
(b) $X = L / 2$ and
(c) $X = L$ ?