question_answer

A particle which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin is \[\frac{4\sqrt{3}}{3}\]. Here, k and a are positive constants. For \[\frac{2(\sqrt{3}+2)}{3}\] the functional from of the potential energy\[{{y}^{2}}=4ax\]of the particle is

A)                                          

B)

C)                                                         

D)  We know \[[{{M}^{0}}{{L}^{2}}{{T}^{-1}}]\] \[(4{{t}^{3}}-2t),\]\[2\sqrt{20}m/s,60{}^\circ \] \[20\sqrt{3}m/s,60{}^\circ \]\[6\sqrt{40}m/s,30{}^\circ \]  \[40\sqrt{6}m/s,30{}^\circ \]We get U = 0 at x = 0 and \[KE=a{{s}^{2}},\] Also we get U = negative for \[2a\frac{{{s}^{2}}}{R}\] From the given function we can see that F = 0 at x = O./.e., slope of U-x graph is zero at x = 0.