Answered

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A force on a particle depends on position such that f(x) = (3. 00 n/m2)x2 + (2. 00 n/m)x for a particle constrained to move along the x-axis. What work is done by this force on a particle that moves from x = 0. 00 m to x = 2. 00 m?.

Sagot :

LammettHash

Integrate the force function over the given displacement:

[tex]W = \displaystyle \int_{x=0.00\,\rm m}^{x=2.00\,\rm m} \left(3.00\frac{\rm N}{\mathrm m^2}\right) x^2 + \left(2.00\frac{\rm N}{\rm m}\right) x \, dx[/tex]

[tex]W = \displaystyle \left(1.00\frac{\rm N}{\mathrm m^2}\right) x^3 + \left(1.00\frac{\rm N}{\rm m}\right) x^2 \bigg|_{x=0.00\,\rm m}^{x=2.00\,\rm m}[/tex]

[tex]W = \displaystyle \left(1.00\frac{\rm N}{\mathrm m^2}\right) (2.00\,\mathrm m)^3 + \left(1.00\frac{\rm N}{\rm m}\right) (2.00\,\mathrm m)^2[/tex]

[tex]W = \displaystyle 8.00 \, \mathrm N{\cdot}\mathrm m + 4.00 \,\mathrm N{\cdot}\mathrm m = \boxed{12.0 \, \mathrm J}[/tex]

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