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Find the work done by the force field f! (x, y) = x sin(y)!i y!j on a particle that moves along the parabola y = x2 from (−1, 1) to (2, 4)

Sagot :

The work done by the force field f! (x, y) = x sin(y)i + yj on a particle that moves along the parabola y = x2 from (−1, 1) to (2, 4) will be  15/2 + 1/2 cos1 -1/2 cos4

Work done on a body is equal to the increase in the energy of the body, for work transfers energy to the body. If, however, the applied force is opposite to the motion of the object, the work is considered to be negative, implying that energy is taken from the object.

The units in which work is expressed are the same as those for energy, for example, in SI (International System of Units) and the metre-kilogram-second system, joule (newton-metre)

Equation of parabola :  y = [tex]x^{2}[/tex]

dy = 2x dx

f(x , y) = x sin (y) i + y j

          =  [tex]\int\limits^2__{} \, -1[/tex] [tex]\ {x sin (x^{2} )} \, dx[/tex] +  [tex]\int\limits^2__{} \, -1[/tex] 2x . [tex]x^{2}[/tex] dx

          =  [tex]\int\limits^2__{} \, -1[/tex] [tex]\ {x sin (x^{2} )} \, dx[/tex] +  [tex]\int\limits^2__{} \, -1[/tex] 2 [tex]x^{3}[/tex] dx

          =   1/2[tex]x^{4}[/tex] * ( - 1/2 cos [tex]x^{2}[/tex] ) + [tex]x^{4}[/tex] / 2

          =   8 - 1/2 + (-1/2 cos 4 + 1/2 cos 1)

          =   15/2 + 1/2 cos1 -1/2 cos4

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