Question

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

Answer 1

The correct option is A $\frac{(a^2+b^2{)}^{1/2}}{m}$

The force in $x$ direction will be the $partial$ derivative of the $-U$ with respect to $y$

so $F_x=-\frac{d}{dx}(ax+by)=-a+0=-a$

The force in $y$ direction will be the $partial$ derivative of the $-U$ with respect to $x$

so $F_y=-\frac{d}{dy}(ax+by)=0-b=-b$

The $net$ force will be $F=\sqrt{F_x^2+F_y^2}=\sqrt{a^2+b^2}$

so the acceleration will be $a=\frac{F}{mass}=\frac{(a^2+b^2{)}^{1/2}}{m}$