Question

A small block with mass 0.0400kg is moving in the xy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.90J/m2 )x2-(3.45J/m3 )y3.

Part A:

What is the magnitude of the acceleration of the block when it is at the point x= 0.21m , y= 0.60m ?

Part B:

What is the direction of the acceleration of the block when it is at the pointx= 0.21m , y= 0.60m ?

Answer 1

The potential energy function is,

$$ \mathrm{U}(x, y)=\left(5.90 \mathrm{~J} / \mathrm{m}^{2}\right) x^{2}-\left(3.45 \mathrm{~J} / \mathrm{m}^{3}\right) y^{3} $$

The force acting along \(\mathrm{x}\) -axis is,

$$ \begin{aligned} F(x) &=\frac{-d U}{d x} \\ &=-\frac{d\left(5.90 \mathrm{~J} / \mathrm{m}^{2}\right) x^{2}}{d x} \\ &=-2(5.90) x \\ &=-11.8 x \end{aligned} $$

The force acting along \(\mathrm{y}\) -axis is,

$$ \begin{aligned} F(y) &=\frac{-d U}{d y} \\ &=-\frac{d\left(-3.45 \mathrm{~J} / \mathrm{m}^{3}\right) y^{3}}{d y} \\ &=-3\left(3.45 \mathrm{~J} / \mathrm{m}^{3}\right) y^{2} \\ &=10.35 y^{2} \end{aligned} $$

The acceleration at \(x\) is,

$$ a_{x}=\frac{F(x)}{m} $$

$$ \begin{array}{l} =\frac{-11.8 x}{m} \\ =\frac{-11.8(0.21 \mathrm{~m})}{0.0400 \mathrm{~kg}} \\ =-61.9 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$

The acceleration at \(y\) is,

$$ \begin{aligned} a_{y} &=\frac{F(y)}{m} \\ &=\frac{10.35 y^{2}}{m} \\ &=\frac{10.35(0.60 \mathrm{~m})^{2}}{0.0400 \mathrm{~kg}} \\ &=93.15 \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

The net acceleration is,

$$ a=\sqrt{a_{n}^{2}+a_{y}^{2}} $$

$$ \begin{array}{l} =\sqrt{\left(-61.9 \mathrm{~m} / \mathrm{s}^{2}\right)^{2}+\left(93.15 \mathrm{~m} / \mathrm{s}^{2}\right)^{2}} \\ =111.8 \mathrm{~m} / \mathrm{s}^{2} \end{array} $$

The direction of acceleration is,

$$ \begin{aligned} \theta &=\tan ^{-1}\left(-\frac{93.15 \mathrm{~m} / \mathrm{s}^{2}}{61.9 \mathrm{~m} / \mathrm{s}^{2}}\right) \\ &=-56.4^{\circ} \end{aligned} $$