In this exercise we want to calculate the distance of a particle from the work it does, so we find the following situation:
[tex]W= 0.128 J[/tex]
Thus, using the work definition we find:
[tex]W= \int\limits{F} \, dx[/tex]
Now replacing the values with the information that was given and the equation, we have:
[tex]W= \int\limits^2_1 {cos(\lambda x/3) } \, dx \\W= \frac{3}{\lambda} [sin(\frac{\lambda x}{3}] \\[/tex]
Applying the points where X is equal to 2 and 1 we find:
[tex]W= \frac{3}{\lambda} [\frac{\sqrt{3} }{2} - \frac{\sqrt{3} }{2}] \\W= 0J[/tex]
The process will be similar for the other points, which correspond respectively to X= 1.5 and 1, in addition to the points X= 2 and 1.5:
[tex]W_2= 0.128 J[/tex]
[tex]W_3= -0.128 J[/tex]
From X=1 to X=1.5, positive work. while X=1.5 to X=2 the work was negative.
See more about work at brainly.com/question/756198
