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  4. A single conservative force F(x) acts on a 1.0- kg particle that moves along the x-axis. The potential energy U(x) is given by U(x)=20+(x-2)2 where x is in meters. At x=5.0 m, the particle has a kinetic energy of 20 J. The maximum kinetic energy of the particle and the value of x at which maximum kinetic energy occurs are

Q. A single conservative force $F(x)$ acts on a $1.0- kg$ particle that moves along the $x$-axis. The potential energy $U(x)$ is given by $U(x)=20+(x-2)^{2}$ where $x$ is in meters. At $x=5.0\, m$, the particle has a kinetic energy of $20\, J$.
The maximum kinetic energy of the particle and the value of $x$ at which maximum kinetic energy occurs are

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Solution:

$KE$ is maximum, where $U$ is minimum and
$U_{\min }=20\, J \text { at } x=2\, m , $
$K_{\max }=E-U_{\min }=49-20=29\, J$

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