A particle moving on $x-axis$ has potential energy, $u = (2 - 20 x + 5 x^{2}) J$ along $x-axis$ . The particle is released at $x = - 3$ .The maximum value of $x$ in $m$ will be

Open in App

Solution

Given, the expression of potential energy, $u = (2 - 20 x + 5 x^{2}) J$
Applying the expression of relation between force and potential energy, we have,

$F = - \frac{d u}{d x}$

$= - \frac{d}{d x} (2 - 20 x + 5 x^{2})$

$= - (0 - 20 + 10 x)$

$\Rightarrow F = - 10 x + 20$

At equilibrium position (mean position), $F = 0$
$\Rightarrow - 10 x + 20 = 0$
$\Rightarrow x = 2 m$
Hence, mean position is at $x = 2 m$

Thus, amplitude of motion, $A = | - 3 | + 2 = 5 m$
Therefore, after reaching $x = 2 m$ , particle will move further by $5 m$ i.e. till $x = 2 + 5 = 7 m$ .
Hence, maximum value of $x$ is $7 m$ .

Accepted answer: 7, 7.0, 7.00

Suggest Corrections

Similar questions

A particle is moving on x-axis has potential energy $U = 2 - 20 x + 5 x^{2}$ J along x-axis. The particle is released at x=-3. The maximum value of x will be [x is in meter and U is in joule]

A particle is moving on x-axis has potential energy $U = 2 - 20 x + 5 x^{2}$ joules along x-axis. The particle is released at x = – 3. If the mass of the particle is 0.1 kg and the maximum velocity of the particle (in m/s) is 10x, the value of x is ___ .