Gujarati
Hindi
5.Work, Energy, Power and Collision
normal

The potential energy of a body of mass $m$ is:
                      $U = ax + by$
Where $x$ and $y$ are position co-ordinates of the particle. The acceleration of the particle is

A

$\frac{{{{({a^2} + {b^2})}^{1/2}}}}{m}$

B

$\frac{{{a^2} + {b^2}}}{m}$

C

$\frac{{{{(a + b)}^{1/2}}}}{m}$

D

$\frac{{a + b}}{m}$

Solution

$F_{x}=-\frac{d v}{d x}=-a$ and $F_{y}=-\frac{d v}{d x}=-b$

Net force, $F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{a^{2}+b^{2}}$

$\text { Acceleration }=\frac{\mathrm{F}}{\mathrm{m}}=\frac{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}{\mathrm{m}}$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.