# The force applied against an object moving horizontally on a linear path is described by F(x)= x^2 +3 N . By how much does the object's kinetic energy change as the object moves from  x in [ 1 , 2 ]?

Jul 28, 2017

$\Delta K = \frac{16}{3} J$

#### Explanation:

Since the object moves on a horizontal linear path, we assume there is no change in potential energy and therefore we can use the work-energy theorem, summarized by the equation $W = \Delta K$.

So, we can calculate the work done on the object and this will be equivalent to the change in kinetic energy.

We know that work can be expressed as the dot (scalar) product of the force and displacement vectors.

$\implies W = F \cdot x \cos \left(\theta\right)$

where $\theta$ is the angle between the vectors. In this case, $\theta = 0$.

Expressing this in terms of differentials, we have:

$\mathrm{dW} = \vec{F} \cdot \mathrm{dv} e c x$

Now we can integrate both sides.

$\implies W = \int F \mathrm{dx}$

Because we are given $x \in \left[1 , 2\right]$ and $F \left(x\right) = {x}^{2} + 3$, we have the definite integral:

${\int}_{1}^{2} \left({x}^{2} + 3\right) \mathrm{dx}$

Integrating, we get $\frac{1}{3} {x}^{3} + 3 x {|}_{1}^{2}$

Which we can evaluate to get $\frac{16}{3}$.

$\therefore \Delta K = \frac{16}{3} J$