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A graph of potential energy V(x) verses x is shown in figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

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प्रश्न

A graph of potential energy V(x) verses x is shown in figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

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उत्तर

KE versus x graph

We know that Total ME = KE + PE

⇒ E0 = KE + V(x)

⇒ KE = E0 – V(x)

At A1x = 0, V(x) = E0

⇒ KE = E0 – E0 = 0

At B1 V(x) < E0

⇒ KE > 0

At C and D1V(x) = 0

⇒ KE is maximum at F1V(x) = E0

Hence, KE = 0

The variation is shown in the adjacent diagram.

Velocity versus x graph

As KE = `1/2` mv2

∴ At A and F, where KE = 0, v = 0

At C and D, KE is maximum.

Therefore, v is ± max.

At B, KE is positive but not maximum.

Therefore, v is ± some value  .....(< max)

The variation is shown in the diagram.

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पाठ 6: Work, Energy and Power - Exercises [पृष्ठ ४६]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 11
पाठ 6 Work, Energy and Power
Exercises | Q 6.34 | पृष्ठ ४६

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