A small block with mass 0.0400kg is moving in the xy -plane. The net force on th
ID: 1375965 • Letter: A
Question
A small block with mass 0.0400kg is moving in thexy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.95J/m2 )x2-(3.55J/m3 )y3.
Part A
What is the magnitude of the acceleration of the block when it is at the point x= 0.21m , y= 0.68m ?
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= 0.21m , y= 0.68m ?
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A small block with mass 0.0400kg is moving in thexy-plane. The net force on the block is described by the potential- energy function U(x,y)= (5.95J/m2 )x2-(3.55J/m3 )y3.
Part A
What is the magnitude of the acceleration of the block when it is at the point x= 0.21m , y= 0.68m ?
Express your answer with the appropriate units.
a =SubmitMy AnswersGive Up
Incorrect; Try Again; 2 attempts remaining
Part B
What is the direction of the acceleration of the block when it is at the point x= 0.21m , y= 0.68m ?
? = ? counterclockwise from the +x-axisSubmitMy AnswersGive Up
Incorrect; Try Again; 4 attempts remaining
Explanation / Answer
U(x,y)= (5.95 )x2-(3.55)y3
x= 0.21m , y= 0.68m
=0.262-1.116
=0.854
workdone=force required to move the block*distance
from the diagram calculate displacement
F=ma
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