Question

A particle is released from rest at origin . its moves under influence of potential field U = x^2-3x ,kinetic energy at x =2

Answer 1

The picture quality is poor but hopefully it is understandable. Cheers! And: (a) 2J

Answer 2

The kinetic energy at x= 2 is 2 Joule.

Solution:

The particle released from the origin which was at rest, under the influence of potential field moves so the kinetic energy can be calculated as follows,

Given: Potential field $U=x^{2}-3 x$  

As we know that the $F=-\frac{d U}{d x}$  

$\begin{array}{l}{F=-(2 x-3)} \\ \\{F=3-2 x} \\ \\{m v \frac{d V}{d x}=3-2 x} \\ \\{m v d v=(3-2 x) d x}\end{array}$

$\begin{array}{l}{m \int v d v=\int(3-2 x) d x} \\ \\{m\left[\frac{v^{2}}{2}\right]=3 x-x^{2}} \\ \\{\frac{1}{2} m v^{2}=3 x-x^{2}} \\ \\{\text {K.E.}=3 x-x^{2}}\end{array}$

Given data suggests that K.E.at x=2

$K . E .=3 \times 2-2 \times 2$

K.E. = 6 – 4

Kinetic Energy = 2  Joule