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A particle-free to move along the X-axis has potential energy given by U(x) = k [1-exp(-x2)] for  where k is a positive constant of appropriate dimensions. Then which of the following are incorrect?
  • a)
    At points away from the origin, the particle is in unstable equilibrium 
  • b)
    For any finite nonzero value of x, there is a force directed away from the origin.
  • c)
    It its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.
  • d)
    For small displacements from x = 0, the motion is simple harmonic.
Correct answer is option 'A,B,C'. Can you explain this answer?
Verified Answer
A particle-free to move along the X-axis has potential energy given by...
Given that U(x) = k[1-e-x2]
The graph of U(x) is shown in fig.(1)

Now F(x) = -
Thus force is zero only at following three points:

At any point away from origin (excluding ) the particle is not even in equilibrium.
Further, at finite nonzero values of x, the force is directed towards origin.


If displaced a little about origin, the particle will execute s.H.M. Again, Total mechanical energy = K.E. + P.E.

or 
The graph of K.E. is also shown in fig. 1 it is obvious from the graph that K.E. is not the minimum at the origin.
 
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Most Upvoted Answer
A particle-free to move along the X-axis has potential energy given by...
Given that U(x) = k[1-e-x2]
The graph of U(x) is shown in fig.(1)

Now F(x) = -
Thus force is zero only at following three points:

At any point away from origin (excluding ) the particle is not even in equilibrium.
Further, at finite nonzero values of x, the force is directed towards origin.


If displaced a little about origin, the particle will execute s.H.M. Again, Total mechanical energy = K.E. + P.E.

or 
The graph of K.E. is also shown in fig. 1 it is obvious from the graph that K.E. is not the minimum at the origin.
 
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A particle-free to move along the X-axis has potential energy given by U(x) = k [1-exp(-x2)] forwhere k is a positive constant of appropriate dimensions. Then which of the following are incorrect?a)At points away from the origin, the particle is in unstable equilibriumb)For any finite nonzero value of x, there is a force directed away from the origin.c)It its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)For small displacements from x = 0, the motion is simple harmonic.Correct answer is option 'A,B,C'. Can you explain this answer?
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A particle-free to move along the X-axis has potential energy given by U(x) = k [1-exp(-x2)] forwhere k is a positive constant of appropriate dimensions. Then which of the following are incorrect?a)At points away from the origin, the particle is in unstable equilibriumb)For any finite nonzero value of x, there is a force directed away from the origin.c)It its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)For small displacements from x = 0, the motion is simple harmonic.Correct answer is option 'A,B,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A particle-free to move along the X-axis has potential energy given by U(x) = k [1-exp(-x2)] forwhere k is a positive constant of appropriate dimensions. Then which of the following are incorrect?a)At points away from the origin, the particle is in unstable equilibriumb)For any finite nonzero value of x, there is a force directed away from the origin.c)It its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)For small displacements from x = 0, the motion is simple harmonic.Correct answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle-free to move along the X-axis has potential energy given by U(x) = k [1-exp(-x2)] forwhere k is a positive constant of appropriate dimensions. Then which of the following are incorrect?a)At points away from the origin, the particle is in unstable equilibriumb)For any finite nonzero value of x, there is a force directed away from the origin.c)It its total mechanical energy is k/2, it has its minimum kinetic energy at the origin.d)For small displacements from x = 0, the motion is simple harmonic.Correct answer is option 'A,B,C'. Can you explain this answer?.
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