Question

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?.

Answer 1

Integrate the force function over the given displacement:

`W = \displaystyle \int_(x=0.00\,\rm m)^(x=2.00\,\rm m) \left(3.00(\rm N)/(\mathrm m^2)\right) x^2 + \left(2.00(\rm N)/(\rm m)\right) x \, dx`

`W = \displaystyle \left(1.00(\rm N)/(\mathrm m^2)\right) x^3 + \left(1.00(\rm N)/(\rm m)\right) x^2 \bigg|_(x=0.00\,\rm m)^(x=2.00\,\rm m)`

`W = \displaystyle \left(1.00(\rm N)/(\mathrm m^2)\right) (2.00\,\mathrm m)^3 + \left(1.00(\rm N)/(\rm m)\right) (2.00\,\mathrm m)^2`

`W = \displaystyle 8.00 \, \mathrm N{\cdot}\mathrm m + 4.00 \,\mathrm N{\cdot}\mathrm m = \boxed{12.0 \, \mathrm J}`