The figure below shows a plot of potential energy (U = PE) versus position (x) o

Question

The figure below shows a plot of potential energy (U = PE) versus position (x) of a 0.200 kg particle that can travel only along an x-axis under the influence of a conservative force. The graph has values UA = 9.00 J, UC = 20.00 J, and UD = 24.00 J. The particle is released at the point where U forms a "potential hill" of "height" UB = 12.00 J, with a velocity of 5.49 m/s in the negative x-direction. What will be the speed of the particle at x = 3.5 m? What is the force on the particle in the region between x = 1.0 m and x = 3.0 m? Will the particle be able to reach x = 0.0 m?

Explanation / Answer

Solution:

mass of the particle = m = 0.200 kg

UA = 9 J ; UC = 20 J and UD = 24 J

UB = 12 J

Velocity of the particle at B = 5.49 m/s

Total energy at B = U+K =12 + 0.5 (0.2)(5.49)2 = 12+3.01= 15 joules

At x = 3.5 m , the potential energy UA = 9 J

Kinetic energy at A , when x = 3.5 m , K = total energy - UA , since total energy is conserved.

1/2m vA2 = 15 - 9 = 6 J => vA2 = 6 * 2 / 0.2 = 60

=> velocity = sqrt (60 ) = 7.746 m/s = 7.75 m/s

Work done = change in potential energy

Force = work done / displacement = UC - UA / x2- x1 = (20 - 9 ) / (3-1) = 11/2 = 5.5 N

The x= 0 position indicates that the particle will be moving along the y direction.

This is not possible Since the particle has enough enrgy to go forward along the x direction itself .

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