JEE MAIN - Physics (2006 - No. 46)
The total mechanical energy of the particle is $$2J.$$ Then, the maximum speed (in $$m/s$$) is
$${3 \over {\sqrt 2 }}$$
$${\sqrt 2 }$$
$${1 \over {\sqrt 2 }}$$
$$2$$
Explanation
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.) =$$ {1 \over 4} - {1 \over 2} = - {1 \over 4}J$$
$$K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,$$ (Given)
$$\therefore$$ $$K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}$$
$$\therefore$$ $${1 \over 2}mv_{\max }^2$$ = $${9 \over 4}$$
$$ \Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}$$
$$ \Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$$ m/s
For minimum potential energy,
$${{dV} \over {dx}} = 0 $$
$$\Rightarrow {x^3} - x = 0 $$
$$\Rightarrow x = \pm 1$$
$$ \Rightarrow$$ Min. Potential energy (P.E.) =$$ {1 \over 4} - {1 \over 2} = - {1 \over 4}J$$
$$K.E{._{\left( {\max .} \right)}} + P.E{._{\left( {\min .} \right)}} = 2\,$$ (Given)
$$\therefore$$ $$K.E{._{\left( {\max .} \right)}} = 2 + {1 \over 4} = {9 \over 4}$$
$$\therefore$$ $${1 \over 2}mv_{\max }^2$$ = $${9 \over 4}$$
$$ \Rightarrow {1 \over 2} \times 1 \times {v^2}_{\max .} = {9 \over 4}$$
$$ \Rightarrow {v_{\max }} = {3 \over {\sqrt 2 }}$$ m/s
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