Question

A one-dimensional variable force in the x-direction is given F (x) =
`2x^(4.79) i` 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.

Answer 1

• 55 joules

Step-by-step explanation:

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:

• 
  `dW=f(x).dx`

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

• 
  `W=\int\limits^(x_f)_(x_i) {f(x)} \, dx`

Thus, for this problem we have:

• xi = 0.73

• xf = 2.4

• f(x) =
  `2x^(4.79)i`

The symbol
`i` is just indicating that the direction of the force is in the same direction of the displacement.

Integrating you get:

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

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.