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Asked Nov 3, 2019
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#11. (10 points) A 150 [g] particle at x 0 is moving at 2.00 [m/s] in the +x-direction
As it moves, it experiences a force given by F (0.250) sin ( What is the
2.00
particle's speed when it reaches r= 3.14 [m]?
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#11. (10 points) A 150 [g] particle at x 0 is moving at 2.00 [m/s] in the +x-direction As it moves, it experiences a force given by F (0.250) sin ( What is the 2.00 particle's speed when it reaches r= 3.14 [m]?

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Expert Answer

Step 1

the work done is equal to the change in kinetic energy,

the work done is, 

Fdx
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Fdx

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Step 2

integrate the above equation by substituting the given values, 

T-
3.14
X
W
0.250sin
dx
2.00
3.14m
(0.250)
-cos
2.00
3.14m
0.500 1-cos
2.00
=0.4996J
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T- 3.14 X W 0.250sin dx 2.00 3.14m (0.250) -cos 2.00 3.14m 0.500 1-cos 2.00 =0.4996J

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Step 3

the change in kinetic ...

AK = m(v-u2)
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AK = m(v-u2)

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