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

Q. A particle free to move along the x-axis has potential energy given by

U (x) = k [1 - e x p (- x^{2})]

for

- \infty \leq x \leq + \infty,

where k is a positive constant of appropriate dimensions. Then:

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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

The correct option is E −14.8NThe potential energy of the particle U(x)=e2x+1Thus force acting on the particle F=−d[U(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

The correct option is E $- 14.8 N$
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
$⟹$ $F {∣ ∣ ∣}_{x = 1} = - 2 e^{2} = - 14.8$ N