# 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).

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A small block with mass

#0.0400 kg# is moving in the xy-plane. The net force on the block is described by the potential- energy function #U(x,y)= (5.90 J/m^2 )x^2-(3.65 J/m^3 )y^3# .

A small block with mass

##### 1 Answer

Since

#vecF=-gradU#

#F_x = -(delU)/(delx)#

#F_x = -del/(delx)[(5.90\ Jm^-2)x^2−(3.65\ Jm^-3)y^3]#

#F_x = -11.80x#

#F_x=ma_x = -11.80x#

#0.0400a_x = -11.80x#

#=>a_x = -11.80/0.0400x#

#=>a_x = -295x#

At the desired point

#a_x = -295xx0.24#

#a_x = -70.8\ ms^-2#

Similarly

#F_y = -del/(dely)[(5.90\ Jm^-2)x^2−(3.65\ Jm^-3)y^3]#

#F_y = 10.95y^2#

#F_y=ma_ = 10.95y^2#

#0.0400a_y = 10.95y^2#

#=>a_y = 10.95/0.0400y^2#

#=>a_y =27.375y^2 #

At the desired point

#a_y = 27.375xx(0.52)^2#

#a_y = 7.4022\ ms^-2#

Now

#|veca| = sqrt[(-70.8)^2 + (7.4022)^2]#

#|veca| =71.2\ ms^-2#

If

#tantheta = (a_y)/(a_x)#

Inserting calculated values

#tantheta = (7.4022)/(-70.8)# , (#2nd# quadrant)

#=>theta=174^@#