# What is the magnitude of the acceleration of the block when it is at the point x= 0.24 m, y= 0.52m? What is the direction of the acceleration of the block when it is at the point x= 0.24m, y= 0.52m? (See details).

## A small block with mass $0.0400 k g$ is moving in the xy-plane. The net force on the block is described by the potential- energy function $U \left(x , y\right) = \left(5.90 \frac{J}{m} ^ 2\right) {x}^{2} - \left(3.65 \frac{J}{m} ^ 3\right) {y}^{3}$.

Feb 19, 2018

Since $x \mathmr{and} y$ are orthogonal to each other these can be treated independently. We also know that

$\vec{F} = - \nabla U$

$\therefore x$-component of two dimensional force is

${F}_{x} = - \frac{\partial U}{\partial x}$
F_x = -del/(delx)[(5.90\ Jm^-2)x^2−(3.65\ Jm^-3)y^3]
${F}_{x} = - 11.80 x$

$x$-component of acceleration

${F}_{x} = m {a}_{x} = - 11.80 x$
$0.0400 {a}_{x} = - 11.80 x$
$\implies {a}_{x} = - \frac{11.80}{0.0400} x$
$\implies {a}_{x} = - 295 x$

At the desired point

${a}_{x} = - 295 \times 0.24$
${a}_{x} = - 70.8 \setminus m {s}^{-} 2$

Similarly $y$-component of force is

F_y = -del/(dely)[(5.90\ Jm^-2)x^2−(3.65\ Jm^-3)y^3]
${F}_{y} = 10.95 {y}^{2}$

$y$-component of acceleration

${F}_{y} = m {a}_{=} 10.95 {y}^{2}$
$0.0400 {a}_{y} = 10.95 {y}^{2}$
$\implies {a}_{y} = \frac{10.95}{0.0400} {y}^{2}$
$\implies {a}_{y} = 27.375 {y}^{2}$

At the desired point

${a}_{y} = 27.375 \times {\left(0.52\right)}^{2}$
${a}_{y} = 7.4022 \setminus m {s}^{-} 2$

Now $| \vec{a} | = \sqrt{{a}_{x}^{2} + {a}_{y}^{2}}$

$| \vec{a} | = \sqrt{{\left(- 70.8\right)}^{2} + {\left(7.4022\right)}^{2}}$
$| \vec{a} | = 71.2 \setminus m {s}^{-} 2$

If $\theta$ is the angle made by acceleration with $x$-axis at the desired point then

$\tan \theta = \frac{{a}_{y}}{{a}_{x}}$

Inserting calculated values

$\tan \theta = \frac{7.4022}{- 70.8}$, ($2 n d$ quadrant)
$\implies \theta = {174}^{\circ}$