Question

The figure above shows the PE curve for a particle interacting with an external field.

1) the particle is released with energy E1 and point B while moving towards the right. Where is the particle traveling the fastest?

2) The particle is released with energy E1 at point B while moving towards the right. WHere is the particle traveling the slowest or stopped?

3) The particle is released with energy E2 at point C while moving towards the right. Where is the particle traveling the fastest?

Answer 1

1. The particle is fastest at point C as its PE is lowest

2. The particle is stopped at point F as its PE is equal to its total energy so KE = 0

3. The particle is moving fastest at point C itself

* maximum slope represent the max kinetic energy where as point C & E represent minimum potential energy (stable condition) where as point B & D represent unstable equilibrium.

Answer 2

1. The particle is fastest at point C as its PE is lowest at that point

2. The particle is stopped at point F as its PE is equal to its total energy, so its KE = 0

3. The particle is moving fastest at point C itself as it does not have enough energy to cross the point D so minimum PE point is point C itself

Answer 3

1) Point C has lowest potential energy so the kinetic energy is maximum. so at point c the particle is fastest

2)potential energy at F is highest so the particle is slowest at that point.

3) the particle is released at point c. and doesnot have enough energy to cross point d. so point c is fastest