Question

A small block with mass 0.0400 kg is moving in thexy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.95 J/m2 )x2-(3.60 J/m3 )y3.

a) What is the magnitude of the acceleration of the block when it is at the point x= 0.30 m , y= 0.56 m?

b)What is the direction of the acceleration of the block when it is at the point x= 0.30 m , y= 0.56 m ?

Answer 1

x componet of force:

$F_{x}=-\frac{dU}{dx}=-2(5.95)x$

acceleration in x:

$F_{x}=-2(5.95)x=ma\rightarrow a=\frac{-2(5.95)0.3}{0.04}$

$a_{x}=-89.25\frac{m}{s^{2}}$

Force in y:

$F_{y}=-\frac{dU}{dy}=3(3.60)y^{2}$

acceleration in y:

$F_{y}=3(3.60)y^{2}=ma\rightarrow a=\frac{3(3.60)0.56^{2}}{0.04}$

$a_{y}=84.672\frac{m}{s^{2}}$

Magnitude of acceleration:

$a=\sqrt{89.25^{2}+84.672^{2}}\rightarrow \boldsymbol{a= 123.02\frac{m}{s^{2}}}$

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direction:

$\tan (\theta )=\frac{a_{y}}{a_{x}}=\frac{84.672}{-89.25}$

$\theta =-43.49^{\circ }$   Meassured Clockwise or

$\theta =316.51^{\circ }$ Meassure counterclockwise