A force acting on a particle moving in the xy plane is given by F = (2yi + x^2 j

Question

A force acting on a particle moving in the xy plane is given by F = (2yi + x^2 j), where F is in newtons and x and y are n meters. The particle moves from the origin to a final position having coordinates x = 4.70 m and y = 4.70 m, as shown in the figure below. (a) Calculate the work done by F on the particle as it moves along the purple path. Enter a number. be integration in two steps, noting that y has a constant value along the first path, and x has a constant value along the second path. J (b) Calculate the work done by F on the particle as it moves along the red path. J (c) Calculate the work done by F on the particle as it moves along the blue path

Explanation / Answer

Let … A --> 1 … B --> 2 … C --> 3 …
… force vector F = ( Fx ) i + ( Fy ) j = ( 2y ) i + ( x ² ) j …
… components … Fx = 2y = Fx ( y ) <-- function of y ...
…………………… Fy = x ² = Fy ( x ) <-- function of x …
Along the path OAC …
… W = W + W = Fx ( 0 ) • dx + Fy ( 4.7 ) • dy …
… Fx ( 0 ) • dx = 0 since Fx ( 0 ) = 0 along OA …
… Fy ( 4.7 ) • dy = Fy ( 4.7 ) dy = 4.7² dy = 4.7² • 4.7 = 103.823...
… W =103.823 …
Along OBC …
… W = W + W = Fy ( 0 ) • dy + Fx ( 4.7) • dx …
… Fy ( 0 ) • dy = 0 since Fy ( 0 ) = 0 along OA …
… Fx ( 4.7) • dx = Fx ( 4.7 ) dx = 2 ( 4.7 ) dx = 8.4 • 4.7 = 39.48 …
… W = 39.48 J …
Along the path OC …
… y ( x ) = m x + b …-->… y ( x ) = m x + b
… y ( 4.7 ) = m • 4.7 = 4.7… so m = 1 … y = x …
… W = Fx • dx + Fy • dy = 2 y • dx + x ² • dy …
……….. = 2 x • d x + y ² • d y = ( 4.7 ) ² + ( 4.7 ) ³ =56.697 J…
The force F is not conservative since W W so the work
done depends on the path taken …

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