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Chapter 7: Problem 37

A particle moves along the \(x\) -axis under the influence of a force \(F=a x^{2}+b,\) where \(a\) and \(b\) are constants. Find its potential energy as a function of position, taking \(U=0\) at \(x=0\)

Short Answer

Expert verified
The potential energy of the particle as a function of position x is \(U(x)=-[(ax^3)/3 + bx]\)

Step by step solution

01

Understanding the relationship between Force and Potential Energy

Potential Energy \(U\) is related to the Force \(F\) by the relation, \(F=-dU/dx\) where \(dU/dx\) represents the derivative of the potential energy with respect to position \(x\). So to find \(U(x)\), we need to integrate \(F(x)\) by reversing the sign and accounting for the constant of integration which is the potential energy at \(x=0\).
02

Setup integral for Potential Energy

Setting up the integration would result in: \(U(x)=-\int F(x)dx\). Substituting \(F=a x^2+b\) into the equation gives \(U(x)=-\int (ax^2+b)dx\)
03

Perform the integration

Now we can perform the integral: the integral of \(x^2\) with respect to \(x\) is \(x^3/3\) and the integral of a constant (\(b\)) with respect to \(x\) is \(bx\). Thus, after integration and reversing the sign we obtain: \(U(x)=-[(ax^3)/3 + bx] + U(0)\)
04

Apply the Initial condition

Applying the condition that at \(x=0\), \(U=0\) (given in the problem), the constant of integration \(U(0)\) is zero. Thus, the potential energy function is \(U(x)=-[(ax^3)/3 + bx]\)

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Most popular questions from this chapter

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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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