A small block with mass 0.0400 kg is moving in the xy -plane. The net force on t

Question

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)= (6.00 J/m2 )x2-(3.60 J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.31 m , y= 0.70 m ?

Express your answer with the appropriate units.

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Part B

What is the direction of the acceleration of the block when it is at the point x= 0.31 m , y= 0.70 m ?

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)= (6.00 J/m2 )x2-(3.60 J/m3 )y3.

Part A

What is the magnitude of the acceleration of the block when it is at the point x= 0.31 m , y= 0.70 m ?

Express your answer with the appropriate units.

a =

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Part B

What is the direction of the acceleration of the block when it is at the point x= 0.31 m , y= 0.70 m ?

=    counterclockwise from the +x-axis

Explanation / Answer

Potential- energy function U(x,y)= (6.00 J/m2 )x^2-(3.60 J/m3 )y^3.

F = -U/x = - 6.0 * (2x)

F = -U/y = 3.60 * (3y^2)

at the point x= 0.31 m , y= 0.70

we have F = -3.72 N

F = 5.292 N

I F(resultant) I = F = [(-3.72)² + (5.292)²] = 6.46 N

since Newton's II law => F = ma, the direction of a is the same as that of F and

magnitude = F / m.

=> I a I = 6.46 / 0.04 = 161.71 m/s^2

Part B :

Direction of a = = tan ¹ [F / F] = 54.89 degrees clockwise from the -tive x-axis (above it) or at (180 - 54.89) = 125.10 degrees anti-clockwise from the +tive x-axis.

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