у (m)31. A 4.00-kg particle movesQIC from the origin to positionT ©, having coordinates x =5.00 m and y(Fig. P7.31). One force onthe particle is the gravita-tional force acting in thenegative y direction. UsingEquation 7.3, calculate thework done by the gravita-tional force on the parti-cle as it goes from O to Calong (a) the purple path, (b) the red path, and (c) the bluepath. (d) Your results should all be identical. Why?(5.00, 5.00)5.00 mx (m)Figure P7.31Problems 31 through 33

Question
Asked Oct 29, 2019
у (m)
31. A 4.00-kg particle moves
QIC from the origin to position
T ©, having coordinates x =
5.00 m and y
(Fig. P7.31). One force on
the particle is the gravita-
tional force acting in the
negative y direction. Using
Equation 7.3, calculate the
work done by the gravita-
tional force on the parti-
cle as it goes from O to C
along (a) the purple path, (b) the red path, and (c) the blue
path. (d) Your results should all be identical. Why?
(5.00, 5.00)
5.00 m
x (m)
Figure P7.31
Problems 31 through 33
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у (m) 31. A 4.00-kg particle moves QIC from the origin to position T ©, having coordinates x = 5.00 m and y (Fig. P7.31). One force on the particle is the gravita- tional force acting in the negative y direction. Using Equation 7.3, calculate the work done by the gravita- tional force on the parti- cle as it goes from O to C along (a) the purple path, (b) the red path, and (c) the blue path. (d) Your results should all be identical. Why? (5.00, 5.00) 5.00 m x (m) Figure P7.31 Problems 31 through 33

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check_circleExpert Solution
Step 1

Part A:

 

The coordinates of point (C) is (5.00 m, 5.00 m) and the mass of the body
is 4.00 kg.
The acceleration due to gravity is 9.8 m/s2
Write the expression for the work done along OA.
WOA mg (OA) cos 90.0
COS
Here, m is the mass of the object and g is the acceleration due to gravity.
Write the expression for the work done along AC
mg (AC) cos 180.0
WAC
Write the expression for the work done along purple path
Wpurple
WOAWAC
Substitute mg (OA) cos 90.0 for WoA and mg (AC) cos 180.0 for WAC in
the above equation.
Wpurple (mg (OA) cos 90.0
+ (mg (AC) cos 180.0*
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The coordinates of point (C) is (5.00 m, 5.00 m) and the mass of the body is 4.00 kg. The acceleration due to gravity is 9.8 m/s2 Write the expression for the work done along OA. WOA mg (OA) cos 90.0 COS Here, m is the mass of the object and g is the acceleration due to gravity. Write the expression for the work done along AC mg (AC) cos 180.0 WAC Write the expression for the work done along purple path Wpurple WOAWAC Substitute mg (OA) cos 90.0 for WoA and mg (AC) cos 180.0 for WAC in the above equation. Wpurple (mg (OA) cos 90.0 + (mg (AC) cos 180.0*

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Step 2
Conclusion:
Substitute 4.00kg for m, 9.8 m/s2 for g, 5.00 m for (AC) and 5.00 m for
(OC) in the above equation
Wpurple
((4.00 kg) 9.8m/s2(5.00 m) cos 90.0
((4.00 kg) 9.8m/s (5.00 m) cos 180.0
0(4.00kg) (9.8m/s2) (5.00 m) (-1)
=-196 J
Therefore, the work done in the purple path is -196 J.
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Conclusion: Substitute 4.00kg for m, 9.8 m/s2 for g, 5.00 m for (AC) and 5.00 m for (OC) in the above equation Wpurple ((4.00 kg) 9.8m/s2(5.00 m) cos 90.0 ((4.00 kg) 9.8m/s (5.00 m) cos 180.0 0(4.00kg) (9.8m/s2) (5.00 m) (-1) =-196 J Therefore, the work done in the purple path is -196 J.

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Step 3

Part B:

 ...

The acceleration due to gravity is 9.8 m/s2.
Write the expression for the work done along BC.
WBс 3 тg (ВC) сos 90.0"
Write the expression for the work done along OB
WOB
= mg (OB) cos 180.0
Write the expression for the work done along red path
Wred WoB WBC
Substitute mg (BC) cos 90.0
for WBC and mg (OB) cos 180.0 for WoB in
the above equation.
(mg (BC) cos 90.0*) (mg (OB) cos 180.0°
Wred
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The acceleration due to gravity is 9.8 m/s2. Write the expression for the work done along BC. WBс 3 тg (ВC) сos 90.0" Write the expression for the work done along OB WOB = mg (OB) cos 180.0 Write the expression for the work done along red path Wred WoB WBC Substitute mg (BC) cos 90.0 for WBC and mg (OB) cos 180.0 for WoB in the above equation. (mg (BC) cos 90.0*) (mg (OB) cos 180.0° Wred

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