First slide

Different Forms of energy and conservation of energy

Question

The potential energy of a 1 kg particle free to move along the x-axis is given by $V (x) = (\frac{x^{4}}{4} - \frac{x 2}{2})$ Joules. The total Mechanical Energy of the particle 2J. Then the maximum speed is,

Moderate

A

2 m/s
B

$\frac{3}{\sqrt{2}} m / s$
C

$\sqrt{2} m / s$
D

$\frac{1}{\sqrt{2}} m / s$

Solution

$\begin{array}{rcl} t o t a l e n e r g y T E & = & 2 J \\ f o r m a x i m u m s p e e d & \Rightarrow & K E i s m a x i m u m, \\ \Rightarrow & P E i s m i n i m u m \\ P E & = & V (x) = \frac{x^{4}}{4} - \frac{x^{2}}{2} \\ d i f f e r e n t i a t e w i t h r e s p e c t t o x, t o f i n d m i n i m u m P E \\ \frac{d V}{d x} & = & \frac{4 x^{3}}{4} - \frac{2 x}{2} \\ \frac{d V}{d x} & = & x^{3} - x \\ f o r m a x i m u m o r m i i n i m u m \frac{d V}{d x} & = & 0 \\ x^{3} - x & = & 0 \\ \Rightarrow & x^{2} = 1 \\ \Rightarrow & x = \pm 1 \\ f o r m i n i m u m v a l u e o f P E c o n s i d e r x & = & - 1 \\ V (x) & = & \frac{x^{4}}{4} - \frac{x^{2}}{2} \\ s u b s t i t u t e x & = & - 1 \\ V (x) & = & - \frac{1}{4} = P E m i n i m u m \\ T E & = & K E + P E \\ K E_{m a x} & = & T E - P E_{m i n} \\ K E_{m a x} & = & 2 - (- \frac{1}{4}) \\ K E_{m a x} & = & \frac{9}{4} \\ \frac{1}{2} m v^{2} & = & \frac{9}{4} \\ v & = & \frac{3}{\sqrt{2}} m / s \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App