Question 184: Work, Energy and Power & Circular Motion

Question

The potential energy of a $$1$$ $$kg$$ particle free to move along the $$x$$-axis is given by $$V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J$$.
The total mechanical energy of the particle is $$2J.$$ Then, the maximum speed (in $$m/s$$) is

Accepted Answer

Velocity is maximum when kinetic energy is maximum and when kinetic energy is maximum then potential energy should be minimum
For minimum potential energy,
$${{dV} \over {dx}} = 0 $$
$$\Rightarrow {x^3} - x = 0 $$
$$\Rightarrow x =  \pm 1$$
$$ \Rightarrow$$ Min. Potential energy (P.E.) =$$ {1 \over 4} - {1 \over 2} =  - {1 \over 4}J$$
$$K.E{._{\left( {\max .} ight)}} + P.E{._{\left( {\min .} ight)}} = 2\,$$ (Given)
$$	herefore$$ $$K.E{._{\left( {\max .} ight)}} = 2 + {1 \over 4} = {9 \over 4}$$
$$	herefore$$ $${1 \over 2}mv_{\max }^2$$ = $${9 \over 4}$$
$$ \Rightarrow {1 \over 2} 	imes 1 	imes {v^2}_{\max .} = {9 \over 4}$$
$$ \Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$$ m/s

Answer

$${3 \over {\sqrt 2 }}$$

Answer

$${\sqrt 2 }$$

Answer

$${1 \over {\sqrt 2 }}$$

Answer

$$2$$

Work, Energy and Power & Circular Motion

The potential energy of a $1 kg particle free to move along the x-axis is given by V\left( x \right) = \left( {{{{x^4}} \over 4} - {{{x^2}} \over 2}} \right)J. The total mechanical energy of the particle is 2J. Then, the maximum speed (in m/s$) is

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