Q.34 A graph of potential energy V (x) versus x is shown in the figure. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.
Step 1: Find KE at different points and plot KE versus x graph.
We know that
Total ME = KE + PE
⇒ E0=KE+V(x)⇒ KE=E0−V(x)
at point A
x=0, V(x)=E0
⇒ KE=E0−E0=0
at point B
V(x)<E0
⇒ KE > 0 (positive)
at points C and D V(x) = 0⇒ KE = E0 (maximum) at point FV(x) = E0
Hence, KE = 0
The variation is shown in the diagram.
Step 2: Use KE=12mv2 and plot velocity vs x graph.
As,
KE=12mv2
∴ At A and F. where KE = 0⇒ v = 0
At C and D, KE is maximum. Therefore, v is ± max.
At B, KE is but not maximum.
Therefore, v is ± some value (< max)
The variation is shown in the diagram.
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