A small block with mass 0.0400 kg is moving in the xy -plane. The net force on t
ID: 1537200 • Letter: A
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 =SubmitMy AnswersGive Up
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-axisExplanation / 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.