Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.

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 :

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]

Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.