Consider a system of particles only one of which is moveable, with the one movea

Question

Consider a system of particles only one of which is moveable, with the one moveable particle confined to move along an x axis. The figure shows a graph of the total potential energy U of this system of particles versus the position of the moveable particle. Five specific x positions, A-E, are marked on the horizontal axis of the graph. The zero level for total potential energy has been chosen so that U(B) and U(D) are both zero. As can be seen in the figure, U(E) > U(C) > U(A). Finally, assume that as the moveable particle is moving, no nonconservative forces are doing work, so that the the total mechanical energy of this system is constant. Assume that the moveable particle is released from rest at position E. Which one of the following statements concerning the subsequent motion of the moveable particle is correct?

Select the one true statement.

1. In comparison with a particle released from rest at position A, the particle released from E will have less kinetic energy when it passes point B. 2. The particle released from E will initially move leftward, but it will never pass point C, so it will never reach point B. 3. The particle released from E will move rightward and never return. 4. In comparison with a particle released from rest at position A, the particle released from E will have more kinetic energy when it passes point B.

Explanation / Answer

Ans:The total energy of the particle is constant always so its Kinetic Energy(KE) + Potential Energy(PE) is always constant.Now its starts from point E where it has only PE.And from the given condition U(E) > U(C) > U(A).And U(D) & U(B) are point of zero potential.So at point D & B the particle has only Kinetic EnergyTowards the right from point E the potential is higher so the particle cannot towards right.Now since U(E) > U(C) so the particle will cross the point C with some finite kinetic energy.The particle will have only kinetic energy at point B, now since U(B)=0 so the kinetic energy of the particle at point B is equal to the potential energy at point E i.e KE(at B)=U(E); Similarly If the particle will release from ponint A then its KE at point B will be U(A). Since U(E) > U(A) So .In comparison with a particle released from rest at position A, the particle released from E will have more kinetic energy when it passes point B. i.e option 4 is correct.

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