A force parallel to the z-axis acts on a particle moving along the z-axis. This force produces a potential energy U(z) given by U(x) = ax¹ where a =1.31 J/m¹ Part A What is the force when the particle is at position a = -0.740 m ? 15. ΑΣΦΑ p ? F₂ = N
A force parallel to the z-axis acts on a particle moving along the z-axis. This force produces a potential energy U(z) given by U(x) = ax¹ where a =1.31 J/m¹ Part A What is the force when the particle is at position a = -0.740 m ? 15. ΑΣΦΑ p ? F₂ = N
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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7.35
![**Exercise 7.35**
A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces a potential energy \( U(x) \) given by \( U(x) = \alpha x^4 \) where \( \alpha = 1.31 \, \text{J/m}^4 \).
### Part A
**Question:**
What is the force when the particle is at position \( x = -0.740 \, \text{m} \)?
**Answer:**
To find the force \( F_x \) at the given position, you need to first understand the relationship between force and potential energy. The force \( F_x \) can be determined using the gradient (or derivative) of the potential energy function:
\[ F_x = -\frac{dU}{dx} \]
Given the potential energy function:
\[ U(x) = \alpha x^4 \]
To find \( F_x \), we take the derivative of \( U(x) \) with respect to \( x \):
\[ \frac{dU}{dx} = 4\alpha x^3 \]
Thus,
\[ F_x = -4\alpha x^3 \]
Substitute \( \alpha = 1.31 \, \text{J/m}^4 \) and \( x = -0.740 \, \text{m} \):
\[ F_x = -4 (1.31) (-0.740)^3 \]
To complete this calculation and find the numerical value of \( F_x \), input these values into the calculation.
#### Graph/Diagram Description:
No graphs or diagrams are depicted in this exercise. The focus is on textual and mathematical analysis.
Click on the 'Submit' button to insert your calculated force value in \( F_x = \text{_____} \, \text{N} \). For additional help or to see the answer, you can click on 'Request Answer'.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fddf14729-dbf2-4574-b793-bebd8df9c378%2F26129ad6-a57a-4dde-be8d-117e164c4216%2Fqkryt5w_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercise 7.35**
A force parallel to the x-axis acts on a particle moving along the x-axis. This force produces a potential energy \( U(x) \) given by \( U(x) = \alpha x^4 \) where \( \alpha = 1.31 \, \text{J/m}^4 \).
### Part A
**Question:**
What is the force when the particle is at position \( x = -0.740 \, \text{m} \)?
**Answer:**
To find the force \( F_x \) at the given position, you need to first understand the relationship between force and potential energy. The force \( F_x \) can be determined using the gradient (or derivative) of the potential energy function:
\[ F_x = -\frac{dU}{dx} \]
Given the potential energy function:
\[ U(x) = \alpha x^4 \]
To find \( F_x \), we take the derivative of \( U(x) \) with respect to \( x \):
\[ \frac{dU}{dx} = 4\alpha x^3 \]
Thus,
\[ F_x = -4\alpha x^3 \]
Substitute \( \alpha = 1.31 \, \text{J/m}^4 \) and \( x = -0.740 \, \text{m} \):
\[ F_x = -4 (1.31) (-0.740)^3 \]
To complete this calculation and find the numerical value of \( F_x \), input these values into the calculation.
#### Graph/Diagram Description:
No graphs or diagrams are depicted in this exercise. The focus is on textual and mathematical analysis.
Click on the 'Submit' button to insert your calculated force value in \( F_x = \text{_____} \, \text{N} \). For additional help or to see the answer, you can click on 'Request Answer'.
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