Exercise 7.37 A small block with mass 0.0400 kg is moving in the xy -plane. The

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Exercise 7.37

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/m2 )x2-(3.85 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= 0.52 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= 0.52 m ?

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Exercise 7.37

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/m2 )x2-(3.85 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= 0.52 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= 0.52 m ?

=    counterclockwise from the +x-axis

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Explanation / Answer

x-component of force:
(Fx) = -U/x
(Fx) = -2(5.90)x
(Fx) = -11.80x

x-component of acceleration:
m(ax) = -11.60x
(0.0400 kg)(ax) = -11.80(0.31m)
(ax) = -91.45m/s²

y-component of force:
(Fy) = -U/y
(Fy) = 3(3.85)y²
(Fy) = 11.55y²

y-component of acceleration:
m(ay) = 11.55y²
(0.0400 kg)(ay) = 11.55(0.52 m)²
(ay) = 150.15 m/s²

a = sqrt[(ax)² + (ay)²]
a = sqrt[(-91.45)² + (150.15)²]

a = 175.8 m/s²

tan = (ay)/(ax)
tan = (150.15)/(-91.45)
= 58.65 °

Required angle is 180-58.65 ° = 121.35 degrees

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