A particle free to move along the x-axis has potential energy given by $U (x) = k (1 - e^{- x^{2}})$ For $- \infty < x < \infty$ , where k is a positive constant of appropriate dimensions. Then

A particle located in a one - dimentional potential field has its potential energy functions as U(x) = $\frac{a}{x^{4}} - \frac{b}{x^{2}}$ , where a and b are positive constants. The position of equilibrium x corresponds to

Join BYJU'S Learning Program

A particle has potential energy dependent on its position on the x axis, represented by the function U(x)=e2x+1 for all real values of x, where U(x) and x are given in standard units. The force it feels at position x=1 is closest to [Take e=2.72]

The correct option is D −14.8 NThe potential energy of the particle U(x)=e2x+1 Thus force acting on the particle F=−dU(x)dx=−2e2x N ⇒F|x=1=−2e2=−14.8 N

A particle has potential energy dependent on its position on the $x$ axis, represented by the function $U (x) = e^{2 x} + 1$ for all real values of $x$ , where $U (x)$ and $x$ are given in standard units. The force it feels at position $x = 1$ is closest to
[Take $e = 2.72$ ]

The correct option is D $- 14.8 N$
The potential energy of the particle $U (x) = e^{2 x} + 1$
Thus force acting on the particle $F = - \frac{d U (x)}{d x} = - 2 e^{2 x} N$
$\Rightarrow {\begin{matrix} F \end{matrix} |}_{x = 1} = - 2 e^{2} = - 14.8 N$