.3 Check Your Understanding Find the work done by the same force in Example 7.4 over a cubic path, y = (0.25 m-2)x³, between the same points A = (0, 0) and B = (2 m, 2 m). F = = (5 N/m)y i +(10 N/m)x j y(m)4 (2, 2) F(x, y) y(x), (0, 0) x(m) Figure 7.6 The parabolic path of a particle acted on by a given force.

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Chapter7: Work And Kinetic Energy
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Find the work done by the force

 

 F→= (5 N/m)x i^ + (10 N/m)y j^

over a cubic path,
y = (0.25 m−2)x^3
 between the same points A = (0, 0) and B = (2 m, 2 m)
,

7.3
Check Your Understanding Find the work done by the same force in Example 7.4 over a cubic path,
y = (0.25 m-2)x° , between the same points A
(0, 0) and B = (2 m, 2 m).
%3D
F = (5 N/m)y i +(10 N/m)x j
y(m)4
(2, 2)
F(x, y)
y(x),
(0, 0)
x(m)
Figure 7.6 The parabolic path of a particle acted on by a
given force.
Transcribed Image Text:7.3 Check Your Understanding Find the work done by the same force in Example 7.4 over a cubic path, y = (0.25 m-2)x° , between the same points A (0, 0) and B = (2 m, 2 m). %3D F = (5 N/m)y i +(10 N/m)x j y(m)4 (2, 2) F(x, y) y(x), (0, 0) x(m) Figure 7.6 The parabolic path of a particle acted on by a given force.
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