Question

The potential energy of a 1 kg particle, free to move along the x-axis, is given by v(x) = j. The total mechanical energy of the potential is 2 j. Then, the maximum speed (in m/s) is

Answer 1

maximum speed of particle is 3/√2 m/s

The potential energy of a 1 kg particle, free to moving along the axis, is given by, V(x) = [x⁴/4 - x²/2] J

here total mechanical energy = 2J

⇒kinetic energy + potential energy = 2J

as total mechanical energy is constant. if maximum kinetic energy( because we have to find maximum speed, kinetic energy will be max), potential energy will be minimum.

let's differentiate V(x) with respect to x,

dV(x)/dx = 4x³/4 - 2x/2 = x³ - x

at dV(x)/dx = 0, x³ - x = 0, x = 0, ± 1

for minimum value of V(x), taking x = ±1

then, minimum potential energy = (1)⁴/4 - (1)²/2 = -1/4 J

now, maximum kinetic energy + ( -1/4 J ) = 2J

⇒maximum kinetic energy = 2 + 1/4 = 9/4 J

⇒1/2 mv_max² = 9/4

⇒1/2 × 1 × v_max² = 9/4

⇒v_max = 3/√2 m/s

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Answer 2

$\huge\bold\purple{Answer:-}$

The potential energy of a 1 kg particle, free to moving along the axis, is given by, V(x) = [x⁴/4 - x²/2] J

here total mechanical energy = 2J

⇒kinetic energy + potential energy = 2J

as total mechanical energy is constant. if maximum kinetic energy( because we have to find maximum speed, kinetic energy will be max), potential energy will be minimum.

let's differentiate V(x) with respect to x,

dV(x)/dx = 4x³/4 - 2x/2 = x³ - x

at dV(x)/dx = 0, x³ - x = 0, x = 0, ± 1

for minimum value of V(x), taking x = ±1

then, minimum potential energy = (1)⁴/4 - (1)²/2 = -1/4 J

now, maximum kinetic energy + ( -1/4 J ) = 2J

⇒maximum kinetic energy = 2 + 1/4 = 9/4 J

⇒1/2 mv_max² = 9/4

⇒1/2 × 1 × v_max² = 9/4

⇒v_max = 3/√2 m/s