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5.work,energy,power and collision — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics5.work,energy,power and collision1 Mark
MCQ
If $F=2 x^2-3 x-2$, then select the correct statement
  • A
    $x=2$ is the position of stable equilibrium
  • B
    $x=-\frac{1}{2}$ is the position of unstable equilibrium
  • ✓
    $x=-\frac{1}{2}$ is the position of stable equilibrium
  • D
    $x=2$ is the position of neutral equilibrium
✓

Answer

Correct option: C.
$x=-\frac{1}{2}$ is the position of stable equilibrium
c
(c)

$F=2 x^2-3 x-2$

Putting $F=0$

$2 x^2-3 x-2=0$

$2 x^2-4 x+x-2=0$

$2 x(x-2)+(x-2)=0$

$(x-2)(2 x+1)=0$

$\Rightarrow x=2, \quad x=\frac{-1}{2}$

$\frac{d^2 v}{d x^2}=\frac{-d F}{d x}=-(4 x-3)$

at $x=\frac{-1}{2}$

$\frac{d^2 v}{d x^2} > 0 \quad \Rightarrow$ Stable equilibrium

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