A particle of mass 100 gm moves in a potential well given by U=8x2-4x+400 joule . Find its acceleration at a distance of 25 cm from equilibrium in the positive direction

Question

A particle of mass 100 gm moves in a potential well given by U=8x2-4x+400 joule . Find its acceleration at a distance of 25 cm from equilibrium in the positive direction (A) 4 m s- 1 (B) 40 m s- 1 (C) -40 m s- 1 (D) -4 m s- 1

Tardigrade Admin · Accepted Answer

We know that F=-(d U/d x) F=-(d/d x)(8 x2 - 4 x + 400) F=-(16 x - 4) F=4-16x At equilibrium 4-16x0=0 and x0=(1/4)m=25cm At the given position x=(25 + 25)=50 cm ⇒ F=4-16× 0.5=-4N or a=(- 4/0.1)=-40ms- 2

Q. A particle of mass $100 g m$ moves in a potential well given by $U = 8 x^{2} - 4 x + 400 j o u l e$ . Find its acceleration at a distance of $25 c m$ from equilibrium in the positive direction

$4 m s^{- 1}$

$40 m s^{- 1}$

$- 40 m s^{- 1}$

$- 4 m s^{- 1}$

Q. A particle of mass 100gm moves in a potential well given by U=8x2−4x+400joule . Find its acceleration at a distance of 25cm from equilibrium in the positive direction

4ms−1

40ms−1

−40ms−1

−4ms−1

We know that F=−dxdU​ F=−dxd​(8x2−4x+400) F=−(16x−4) F=4−16x At equilibrium 4−16x0​=0 and x0​=41​m=25cm At the given position x=(25+25)=50cm ⇒ F=4−16×0.5=−4N or a=0.1−4​=−40ms−2

Q. A particle of mass $100 g m$ moves in a potential well given by $U = 8 x^{2} - 4 x + 400 j o u l e$ . Find its acceleration at a distance of $25 c m$ from equilibrium in the positive direction

$4 m s^{- 1}$

$40 m s^{- 1}$

$- 40 m s^{- 1}$

$- 4 m s^{- 1}$