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Marrrta [24]
3 years ago
11

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/m

3)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.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

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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2 years ago
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The answer will be 936 N.

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Given that

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The answer will be 936 N.

                                                               

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