Q. 61 The potential energy for a parti... [FREE SOLUTION]

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Chapter 10: Q. 61 (page 259)

The potential energy for a particle that can move along the x-axis is $U = A x^{2} + B \sin (\frac{蟺 x}{L})$ 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

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

Step by step solution

Given information (part a)

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

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

Explanation (part a)

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

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

At $X = 0$

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

Given information (part b)

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

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

Explanation (part b)

$A t X = L / 2$

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

Given information (part c)

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

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

Explanation (part c)

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

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91影视

The potential energy for a particle that can move along the x-axis is $U = A x^{2} + B \sin (\frac{蟺 x}{L})$ 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)

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91影视

The potential energy for a particle that can move along the x-axis is U=Ax2+Bsin(蟺xL)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

Oscillations

Work Energy and Power

Fluids

Geometrical and Physical Optics

Energy Physics

Scientific Method Physics

Study anywhere. Anytime. Across all devices.

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