Question

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 as $F ( x ) = - k x + a x ^ { 3 }$. Here k and a are positive constants. For $x \geq 0$ , the functional from of the potential energy $U _ { ( x ) }$ of the particle is

• A.
• B.
• C.
• D.

Answer 1

Answer: (D)

Answer 2

$F = - \frac { d U } { d x }$ ⇒ $d U = - F . d x$ ⇒ $U = - \int _ { 0 } ^ { x } \left( - k x + a x ^ { 3 } \right) d x$

⇒ $U = \frac { k x ^ { 2 } } { 2 } - \frac { a x ^ { 4 } } { 4 }$

∴ We get $x = \sqrt { \frac { 2 k } { a } }$

Also we get $U =$ negative for $x > \sqrt { \frac { 2 k } { a } }$

From the given function we can see that F = 0 at x = 0 i.e. slope of U-x graph is zero at x = 0.