Question: The given plot shows the variation of U, the potential energy of interaction between two particles with the distance separating them r.

1. B and D are equilibrium points

2. C is a point of stable equilibrium

3. The force of interaction between the two particles is attractive between points C and D and repulsive between D and E

4. The forces of interaction between particles is repulsive between

points E and F.

Which of the above statements are correct?

(a) 1 and 2                                               (b) 1 and 4

(c) 2 and 4                                               (d) 2 and 3

Solution: At C, potential energy is minimum. So, it is stable equilibrium position.



Negative force means attraction and positive force means repulsion.

Question: The potential energy f in joule of particle of mass 1 kg moving in x-y plane obeys the law, f = 3x + 4y. Here, x and y are in metres. If the particle is at rest at (6m, 8m) at time 0, then the work done by conservations force on the particle from the initial position to the instant when it crosses the x-axis is

(a) 25 J                                                    (b) -25 J

(c) 50 J                                                    (d) -50 J



Since, particle was initially at rest. So, it will move in the direction of force.

We can see that initial velocity is in the direction of PO. So the particle will cross the X-axis at origin.

Ki + Ui = Kf + Uf

   0 + ( 3 x 6 + 4 x 8) = Kf + (3 x 0+ 4 x 0)

or    Kf = 50 J                                  Answer (c)