The potential energy in joules of a particle | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$U=3 x+4 y$

$a_{y}=\frac{F_{y}}{m}=\frac{-(\partial U / \partial x)}{m}=-3$

$a_{x}=\frac{F_{y}}{m}=\frac{-(\partial U / \partial y)}{m}=-4$

$

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

$U=3 x+4 y$

$a_{y}=\frac{F_{y}}{m}=\frac{-(\partial U / \partial x)}{m}=-3$

$a_{x}=\frac{F_{y}}{m}=\frac{-(\partial U / \partial y)}{m}=-4$

$\Rightarrow|\overrightarrow{ a }|=5 m / s ^{2}$

Let at time 't' particle crosses y-axis

then $\quad-6=\frac{1}{2}(-3) t^{2}$

$\Rightarrow \quad t=2$ sec.

Along y-direction:

$\Delta y=1 / 2(-4)(B)^{2}=-8$

$\Rightarrow$ particle crosses y-axis at $y=-4$

$A t(6,4): U=34 \& K E=0$

$A t(0,-4): U=-16 \Rightarrow K E=50$

or, $1 / 2 mv ^{2}=50$

$\Rightarrow v =10 m / s$ while crossing $y$ -axis

The potential energy in joules of a particle of mass $1\, kg$ moving in a plane is given by $U = 3x + 4y$, the position coordinates of the point being $x$ and $y$, measured in metres. If the particle is initially at rest at $(6,4)$, then

Similar Questions

A block of mass $1\,kg$ slides on a rough horizontal surface. If the speed of the block decreases from $10\,m/s$ to $8\,m/s$ , the thermal energy developed in this process .................. $\mathrm{J}$

A ball of mass $0.2\ kg$ is thrown vertically upwards by applying a force by hand. If the hand moves $0.2\ m$ while applying the force and the ball goes upto $2\ m$ height further, find the magnitude of the force .......... $N$ .(consider $g = 10\; m/s^2$)