Question

A particle of mass $1$ kg has potential energy, P. E. $=$ $3x+4y$. At $t$ $=0$, the particle is at rest at $(6,4)$. Work done by external force to displace the particle from rest to the point of crossing the x axis is :

• A. -25 J
• B. 20 J
• C. 15 J
• D. 52 J

Answer 1

Answer: (A)

-25 J

$F_x=-\frac{d(P.E)}{dx}=-3$ and $F_y=-\frac{d(P.E)}{dy}=-4$
$a_x=\frac{F_x}{m}=\frac{-3}{1}=-3$ and $a_y=\frac{F_y}{m}=\frac{-4}{1}=-4$

Since $a_x$ and $a_y$ are constant, we can apply the equations of kinematics.
The particle is at rest at (6,4) at $t=0$

$\therefore$ $u_x=0$ and $u_y=0$

Using the one dimensional equation for kinematics:

$s\left(t\right)=s\left(0\right)+ut+\frac{1}{2}a{t}^2$
$\therefore x=6+\frac{1}{2}{a}_x{t}^2=6-\frac{3}{2}{t}^2$
and $y=4+\frac{1}{2}{a}_y{t}^2=4-\frac{4}{2}{t}^2$

$a=\sqrt{a_x^2+a_y^2}=(3^2+4^2{)}^{1/2}=5m{s}^{-2}$

Now $y=0$ when particle crosses x-axis

$\Rightarrow 4-\frac{4}{2}{t}^2=0$

$\Rightarrow t=\sqrt{2}$
Displacement $S=\frac{1}{2}a{t}^2=\frac{1}{2}\times 5\times (\sqrt{2}{)}^2=5$

Work done: $W=\overrightarrow{F}\cdot \overrightarrow{s}=(-5)\times 5=-25J$