A particle is placed at the origin and a force F = kx is acting on it (where, k is a positive constant). If U (0) = 0 , the graph of U (x ) versus X will be, (where , U is the potential energy function)

Question

A particle is placed at the origin and a force F = kx is acting on it (where, k is a positive constant). If U (0) = 0 , the graph of U (x ) versus X will be, (where , U is the potential energy function) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

From, F=- (d U/d x) d U =-F d x U =∫0u(x) d U=-∫0x F d x =-∫0x k x d x U =(k x2/2) As, U(0)=0, α x2 (where α is a constant) and U are negative.

Q. A particle is placed at the origin and a force $F = k x$ is acting on it (where, $k$ is a positive constant). If $U (0) = 0$ , the graph of $U (x)$ versus $X$ will be, (where , $U$ is the potential energy function)

From, $F = - \frac{d U}{d x}$
$d U = - F d x$
$U = \int_{0}^{u (x)} d U = - \int_{0}^{x} F d x$
$= - \int_{0}^{x} k x d x$
$U = \frac{k x ^{2}}{2}$
As, $U (0) = 0, α x^{2}$ (where $α$ is a constant) and $U$ are negative.

Q. A particle is placed at the origin and a force F=kx is acting on it (where, k is a positive constant). If U(0)=0, the graph of U(x) versus X will be, (where , U is the potential energy function)

From, F=−dxdU​ dU=−Fdx U=∫0u(x)​dU=−∫0x​Fdx =−∫0x​kxdx U=2kx2​ As, U(0)=0,αx2 (where α is a constant) and U are negative.

Q. A particle is placed at the origin and a force $F = k x$ is acting on it (where, $k$ is a positive constant). If $U (0) = 0$ , the graph of $U (x)$ versus $X$ will be, (where , $U$ is the potential energy function)

From, $F = - \frac{d U}{d x}$
$d U = - F d x$
$U = \int_{0}^{u (x)} d U = - \int_{0}^{x} F d x$
$= - \int_{0}^{x} k x d x$
$U = \frac{k x ^{2}}{2}$
As, $U (0) = 0, α x^{2}$ (where $α$ is a constant) and $U$ are negative.