A one-dimensional variable force in the x-direction is given F (x) =[tex]2x^{4.79} i[/tex] where F is in newtons and x is in meters. Find the work done by this force for taking an object of mass m=2.9kg from xi = 0.73 to xf =2.4.

Work is the change in kinetic energy and may be calculated as the product of the force in the direction of the displacement times the displacement.

For a differential displacement, Δx, and a variable force, f(x), the differential work done is:

[tex]dW=f(x).dx[/tex]

And the total work done from a point xi to xf is:

[tex]W=\int\limits^{x_f}_{x_i} {f(x)} \, dx[/tex]

Thus, for this problem we have:

xi = 0.73

xf = 2.4

f(x) = [tex]2x^{4.79}i[/tex]

The symbol [tex]i[/tex] is just indicating that the direction of the force is in the same direction of the displacement.

Integrating you get:

[tex]W=\int\limits^{x_f}_{x_i} {f(x)} \, dx=\int\limits^{2.4}_{0.73} {2x^{4.79} \, dx=2\times (1/5.79)\times (2.4^{5.79}-0.73^{5.79})[/tex]

And that is 54.8697 joules (since the units for x are meter and the units for f(x) are newtons).

Rounded to two significant digits: 55 joules.