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Chapter 7: Problem 27

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 \((\mathrm{c}) x=-1.4 \mathrm{m}\)

Short Answer

Expert verified
The forces on the particle at \(x = 2.1\) m, \(x = 0\), and \(x = -1.4\) m are \(-67.2\) N, \(0\) N, and \(44.8\) N, respectively.

Step by step solution

01

Physical Concepts and Formulas

The key physical concept here pertains to how potential energy impacts the force experienced by a particle. The force experienced by a particle is the derivative of the potential energy U with respect to the position x. The formula for force is given by \(F = - \frac{dU}{dx}\). This equation gives us the force as a function of position and potential energy.
02

Deriving the force function

Applying the formula, the derivative of the potential energy function \(U(x)=16 x^{2}-4.0\) with respect to \(x\) is needed. The derivative is \(32x\). Therefore, the force \(F\) acting on the particle can be expressed as \(F = -32x\).
03

Calculating the force at \(x=2.1 \mathrm{m}\)

Substitute \(x=2.1\) m into the force function to find the force at this point: \(F = -32*2.1 = -67.2\) N.
04

Calculating the force at \(x=0\)

Substitute \(x=0\) into the force function to find the force at this point: \(F =-32*0 =0\) N.
05

Calculating the force at \(x=-1.4 \mathrm{m}\)

Substitute \(x=-1.4\) m into the force function to find the force at this point: \(F = -32*-1.4 = 44.8\) N.

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Most popular questions from this chapter

A biophysicist grabs the ends of a DNA strand with optical tweezers and stretches it \(26 \mu \mathrm{m}\). How much energy is stored in the stretched molecule if its spring constant is \(0.046 \mathrm{pN} / \mu \mathrm{m} ?\)

(a) Derive an expression for the potential energy of an ob subject to a force \(F_{x}=a x-b x^{3},\) where \(a=5 \mathrm{N} / \mathrm{m}\) \(b=2 \mathrm{N} / \mathrm{m}^{3},\) taking \(U=0\) at \(x=0 .\) (b) Graph the poten energy curve for \(x>0\) and use it to find the turning points an object whose total energy is \(-1 \mathrm{J}\)

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 spring of constant \(k=340 \mathrm{N} / \mathrm{m}\) is used to launch a \(1.5-\mathrm{kg}\) block along a horizontal surface whose coefficient of sliding friction is \(0.27 .\) If the spring is compressed \(18 \mathrm{cm},\) how far does the block slide?

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