Q. 34

# A graph of potential energy V (x) verses x is shown in Fig. 6.12. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

i.) Kinetic energy vs x

We know that,

At C, P.E =0,

Till point D there is no change in P.E, the K.E remains constant

At F,

From B to C the potential decreases so the kinetic energy increases we get inverted curve of potential.

From C to D potential remains constant so the kinetic energy will also remain same and thus we get a straight line parallel to x- axis

From D to F the potential energy increases to Eo linearly so the kinetic energy must fall linearly to Eo.

we get the following plot,

ii.) Velocity vs X

Now,

-----(2)

From B to C, from above equation, velocity increases parabolically symmetric from x-axis.

From C to D there is no change in P.E energy so the velocity remains constant.

From D to F, the P.E increases linearly and thus the velocity decreases parabolically from equation (2)

Now since we get two values of velocities from equation (2), we will get a mirror image of the graph in the negative side of the fourth quadrant too.

we obtain following the graph,

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