The force applied against an object moving horizontally on a linear path is described by F(x)= 1/x . By how much does the object's kinetic energy change as the object moves from  x in [ 1, 3 ]?

Feb 27, 2016

$\delta K = \log 3$

Explanation:

The force experienced by the object as it moves along the x-axis is $F \left(x\right) = \frac{1}{x}$ at each point in the x-axis

Now, work done on an object is equal to the change in kinetic energy of the object, i.e $\delta K = \delta W$. But $\delta W = \setminus {\int}_{a}^{b} \setminus \vec{F} \left(x\right) \setminus \cdot \setminus \hat{\mathrm{dx}}$

So, we first find work done on the object, since the direction of force and movement are in the same direction, $\setminus \vec{F} \setminus \cdot \setminus \hat{\mathrm{dx}} = F \mathrm{dx}$

So, ${\int}_{1}^{3} \left(\frac{1}{x} \cdot \mathrm{dx}\right) = {\left[\log x\right]}_{1}^{3}$

Applying the limits gives you the work done on the object, and since work done on the object is the change in kinetic energy of that given object, the change in kinetic energy is as given in the answer.