Question

A particle is released from rest at origin. It moves
under the influence of potential field U = x2 – 3x
Its kinetic energy at x = 2 is
(1) 2 J
(2) 1.5J
(3) 15
(4) Zero​

Answer 1

Answer:

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 xU=x2−3x  

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

\begin{lgathered}\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}\end{lgathered}F=−(2x−3)F=3−2xmvdxdV=3−2xmvdv=(3−2x)dx

\begin{lgathered}\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}\end{lgathered}m∫vdv=∫(3−2x)dxm[2v2]=3x−x221mv2=3x−x2K.E.=3x−x2

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

K . E .=3 \times 2-2 \times 2K.E.=3×2−2×2

K.E. = 6 – 4

Kinetic Energy = 2  Joule