Question

potential energy of particle of mass 2kg oscillating on x axis is given as U = (x-2)^2 -10. the total energy of oscillation is 26J find maximum speed of particle.
1) 4m/s
2)8m/s
3)√13m/s
4)6m/s​

Answer 1

The maximum speed of the particle is 6 m/s.

(4) is correct option.

Explanation:

Given that,

Mass of particle = 2 kg

Total energy = 26 J

Given equation of potential energy

$U=(x-2)^2-10$

$U=x^2+4-4x-10$

$U=x^2-4x-6$

On differentiating

$\dfrac{dU}{dt}=2x-4$

For maximum speed the kinetic energy should be maximum

$\dfrac{dU}{dt}=0$

So,

$x^2-4=0$

$x=2$

Put the value in the given equation

$U=2^2-4\times2-6$

$U=4-8-6=-10\ J$

We need to calculate the maximum speed of the particle

Using formula of total energy

$T.E=U+K$

Where, U = potential energy

K=kinetic energy

Put the value into the formula

$26=-10+\dfrac{1}{2}mv^2$

$26=-10+\dfrac{1}{2}\times2\times v^2$

$v^2=36$

$v=6\ m/s$

Hence, The maximum speed of the particle is 6 m/s.

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Topic : Potential energy

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Answer 2

Explanation:

Given that,

Mass of particle = 2 kg

Total energy = 26 J

Given equation of potential energy

U=(x-2)^2-10U=(x−2)2−10

U=x^2+4-4x-10U=x2+4−4x−10

U=x^2-4x-6U=x2−4x−6

On differentiating

\dfrac{dU}{dt}=2x-4dtdU=2x−4

For maximum speed the kinetic energy should be maximum

\dfrac{dU}{dt}=0dtdU=0

So,

x^2-4=0x2−4=0

x=2x=2

Put the value in the given equation

U=2^2-4\times2-6U=22−4×2−6

U=4-8-6=-10\ JU=4−8−6=−10 J

We need to calculate the maximum speed of the particle

Using formula of total energy

T.E=U+KT.E=U+K

Where, U = potential energy

K=kinetic energy

Put the value into the formula

26=-10+\dfrac{1}{2}mv^226=−10+21mv2

26=-10+\dfrac{1}{2}\times2\times v^226=−10+21×2×v2

v^2=36v2=36

v=6\ m/sv=6 m/s

Hence, The maximum speed of the particle is 6 m/s.