Question

The potential energy function of a particle due to some gravitational field is given by $U=6x+4y$. The mass of the particle is $1kg$ and no other force is acting on the particle. The particle was initially at rest at a point $(6,8)$. Then the time (in sec.) after which this particle is going to cross the $X$ axis would be:

Answer 1

$F=-\frac{dU}{dx}\hat{i}-\frac{dU}{dy}\hat{j}=-6\hat{i}-4\hat{j}$

$a_x=6m/s^2$

$a_y=4m/s^2$, mass of the particle being $1kg$.

Time at which it crosses $X$ axis will be when y coordinate becomes zero:

$y\left(t\right)=y_{\circ }-\frac{1}{2}{a}_y{t}^2$

or, $0=8-\frac{1}{2}\times 4\times t^2$

$\Rightarrow t=2s$