The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m3)x3−(15J/m2

Question

The potential energy for a certain mass moving in one dimension is given by U(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x)=(2.0J/m3)x3−(15J/m2)x2+(36J/m)x−23JU(x) = (2.0 {\rm J/m}^{3})x^{3}- (15 {\rm J/m}^{2})x^{2}+ (36 {\rm J/m})x – 23 {\rm J}. Find the location(s) where the force on the mass is zero.

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Eirian 5 years 2021-09-04T15:06:44+00:00 1 Answers 28 views 0

Answers ( )

  1. thaiduong
    0
    2021-09-04T15:08:23+00:00
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    Answer:x=2 and x=3

    Explanation:

    Given

    Potential Energy for a certain mass is

    U(x)=2x^3-15x^2+36x-23

    and we know force is given by

    F=-\frac{\mathrm{d} U}{\mathrm{d} x}

    F=-(2\times 3x^2-15\times 2x+36)

    For Force to be zero F=0

    \Rightarrow 6x^2-30x+36=0

    \Rightarrow x^2-5x+6=0

    \Rightarrow x^2-2x-3x+6=0

    \Rightarrow (x-2)(x-3)=0

    Therefore at x=2 and x=3 Force on particle is zero.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )