# Question #e4198

Mar 29, 2017

We know that the conservative force $\vec{F}$ is related to the potential energy function U(x,y) in two dimension through the following relationship

$\vec{F} = - \frac{\partial U}{\partial x} \hat{i} - \frac{\partial U}{\partial y} \hat{j}$

Here it is given that $U = 6 x + 6 y$

So

$\vec{F} = - \frac{\partial \left(6 x + 8 y\right)}{\partial x} \hat{i} - \frac{\partial \left(6 x + 8 y\right)}{\partial y} \hat{j}$

$\implies \vec{F} = \left(- 6 \hat{i} - 8 \hat{j}\right) N$

So acceleration $\vec{a} = \frac{\vec{F}}{M}$

where $M =$ mass of the body moving $= 2 k g$

Hence $\vec{a} = \frac{\vec{F}}{2} = \left(- 3 \hat{i} - 4 \hat{j}\right) m \text{/} {s}^{2}$

When the body crosses the Y-axis, then x=0.
Again it is given that body is at rest at the position (6m,4m) at t=0
This means
at x=6m, velocity =0;
When the body crosses Y-axis,its displacement along X-axis becomes ${s}_{x} = - 6 m$.

If this displacement occurs in t sec due to its movement with acceleration along X axis ${a}_{x} = - 3 m \text{/} {s}^{2}$

then we can write

${s}_{x} = \frac{1}{2} {a}_{x} {t}^{2}$

$\implies - 6 = \frac{1}{2} \times \left(- 3\right) {t}^{2}$

$\implies t = 2 s$