Ncert Exercises & Solutions

Q.34 A graph of potential energy V (x) versus x is shown in the figure. A particle of energy $E_{0}$ $E_{0}$ is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.

Hint: Use total mechanical energy,

E_{0} = KE + V (x) .

$E_{0} = KE + V (x) .$

Step 1: Find KE at different points and plot KE versus x graph.
We know that

Total ME = KE + PE

$\begin{array}{l} \Rightarrow E_{0} = KE + V (x) \\ \Rightarrow KE = E_{0} - V (x) \end{array}$

at point A

$x = 0, V (x) = E_{0}$

$\Rightarrow KE = E_{0} - E_{0} = 0$

at point B

$V (x) < E_{0}$

$\Rightarrow KE > 0 (positive)$

$at points C and D V (x) = 0 \Rightarrow KE = E_{0} (maximum) at point F V (x) = E_{0}$

Hence, KE = 0

The variation is shown in the diagram.

Step 2: Use $K E = \frac{1}{2} m v^{2}$ and plot velocity vs x graph.

As,

$K E = \frac{1}{2} m v^{2}$

$∴$ At A and F. where KE = 0 $\Rightarrow$ 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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