The potential energy of a particle of mass 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The potential energy is given by $V(x) = \frac{x^4}{4} - \frac{x^2}{2}$.To find the equilibrium points,we set the derivative to zero: $\frac{dV}{d

Vedclass · Accepted Answer

$3/\sqrt{2}$

Vedclass · Answer

(A) The potential energy is given by $V(x) = \frac{x^4}{4} - \frac{x^2}{2}$.To find the equilibrium points,we set the derivative to zero: $\frac{dV}{dx} = x^3 - x = 0$.This gives $x(x^2 - 1) = 0$,so $x = 0, 1, -1$.The potential energy is minimum at $x = \pm 1$.$V_{\min} = V(\pm 1) = \frac{(\pm 1)^4}{4} - \frac{(\pm 1)^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.Since the total mechanical energy $E = K.E. + V$ is constant at $2 \ J$,the kinetic energy is maximum when potential energy is minimum.$K.E._{\max} = E - V_{\min} = 2 - (-1/4) = 2 + 0.25 = 2.25 \ J$.Using $K.E._{\max} = \frac{1}{2} m v_{\max}^2$ with $m = 1 \ kg$:$\frac{1}{2} 	imes 1 	imes v_{\max}^2 = \frac{9}{4}$.$v_{\max}^2 = \frac{9}{2}$.$v_{\max} = \frac{3}{\sqrt{2}} \ m/s$.

The potential energy of a particle of mass $1 \ kg$ moving freely along the $x$-axis is given by $V(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. If the total mechanical energy of the particle is $2 \ J$,what is its maximum speed in $m/s$?

Similar Questions

While a person climbs stairs,the gravitational potential energy of the person increases. The source of this energy is

When an arrow is released from a bow,from where does the arrow acquire its kinetic energy?

The potential energy of a particle oscillating along the $x-$axis is given by $U = 20 + (x - 2)^2$,where $U$ is in joules and $x$ is in meters. The total mechanical energy of the particle is $36\,J$. The maximum kinetic energy of the particle is ................ $J$.

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity,the work required to pull the hanging part onto the table is:

$A$ body of mass $2\, kg$ is situated at a height $5\, m$ above the ground. Its potential energy is reported as $70\, J$. The potential energy of a $3\, kg$ body situated at a height $1\, m$ above the ground will be reported as ................ $J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module