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Chapter 7: Problem 36

The force on a particle is given by \(\vec{F}=A \hat{i} / x^{2},\) where \(A\) is a positive constant. (a) Find the potential-energy difference between two points \(x_{1}\) and \(x_{2},\) where \(x_{1}>x_{2}\). (b) Show that the potentialenergy difference remains finite even when \(x_{1} \rightarrow \infty\)

Short Answer

Expert verified
Hence, the potential-energy difference, \(U(x_{2}) - U(x_{1})\), equals to \(A [(1/x_{2}) - (1/x_{1})]\). Even when \(x_{1} \rightarrow \infty\), this difference remains finite and equals \(A/x_{2}\).

Step by step solution

01

Convert Force to Potential Energy

We know that the force is related to potential energy through the relation: \[ \vec{F} = - \frac{dU(x)}{dx} \]. We are given \(\vec{F} = A \hat{i} / x^{2}\), implying: \[- \frac{dU(x)}{dx} = A / x^{2}\]. Note that we ignore the directional unit vector \(\hat{i}\) as we're focused on the magnitude.
02

Integrate to find Potential Energy

We can rearrange and integrate this equation to find the differential potential energy: \[dU(x) = - A dx / x^{2}\], The integral will be: \[U(x) = -A \int dx / x^{2}\], which yields \(U(x) = A/x\) with the help of elementary calculus.
03

Calculate Energy Difference

To find the potential energy difference between two points, \(U(x_{2})\) and \(U(x_{1})\), we subtract the potential energies: \(U(x_{2}) - U(x_{1}) = A/x_{2} - A/x_{1} = A [(1/x_{2}) - (1/x_{1})]\).
04

Analyze Limit as \(x_{1} \rightarrow \infty\)

To show that the potential-energy difference remains finite even when \(x_{1} \rightarrow \infty\), we take the limit: \[ \lim_{x_{1} \rightarrow \infty} A [(1/x_{2}) - (1/x_{1})] = A (1/x_{2} - 0) = A/x_{2} \]. This is a finite value, which confirms the hypothesis.

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Most popular questions from this chapter

A particle of mass \(m\) is subject to a force \(\vec{F}=(a \sqrt{x}) \hat{\imath},\) where \(a\) is a constant. The particle is initially at rest at the origin and is given a slight nudge in the positive \(x\) -direction. Find an expression for its speed as a function of position \(x\)

A particle is trapped in a potential well described by \(U(x)=16 x^{2}-b,\) with \(U\) in joules, \(x\) in meters, and \(b=4.0 \mathrm{J}\) Find the force on the particle when it's at (a) \(x=2.1 \mathrm{m}\) (b) \(x=0,\) and \((\mathrm{c}) x=-1.4 \mathrm{m}\)

Find the potential energy of a 70 -kg hiker (a) atop New Hampshire's Mount Washington, \(1900 \mathrm{m}\) above sea level, and (b) in Death Valley, California, 86 m below sea level. Take the zero of potential energy at sea level.

A \(60-\mathrm{kg}\) hiker ascending \(1250-\mathrm{m}\) -high Camel's Hump mountain in Vermont has potential energy \(-240 \mathrm{kJ} ;\) the zero of potential energy is taken at the mountaintop. What's her altitude?

If the potential energy is zero at a given point, must the force also be zero at that point? Give an example.
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