# The force applied against an object moving horizontally on a linear path is described by F(x)= 4x+3 N . By how much does the object's kinetic energy change as the object moves from  x in [ 1 , 3 ]?

Apr 18, 2017

Given force applied against an object moving horizontally on a linear path
$F \left(x\right) = 4 x + 3$ ......(1)
where it is assumed $x$ is in meters

Since the object moves horizontally on a linear path, we can assume that potential energy of the object does not change. As such only change is in its kinetic energy due to work done against the force.

We know that work done $\mathrm{dW}$ by force while moving through distance $\mathrm{dx}$ is
$\mathrm{dW} = \vec{F} \cdot \mathrm{dv} e c x$
Since both force and direction of motion are co-linear angle between the two is $= {0}^{\circ} \mathmr{and} \therefore \cos {0}^{\circ} = 1$
$\therefore \mathrm{dW} = F \mathrm{dx}$

As the force moves through distance $x \in \left[1 , 3\right]$, total work done against the object or change in its kinetic energy
$W = {\int}_{1}^{3} F \mathrm{dx}$
Inserting value of force from (1) we have
$W = {\int}_{1}^{3} \left(4 x + 3\right) \mathrm{dx}$
$\implies W = {\left[4 {x}^{2} / 2 + 3 x\right]}_{1}^{3}$
$\implies W = \left[\left(2 \times {3}^{2} + 3 \times 3\right) - \left(2 \times {1}^{2} + 3 \times 1\right)\right]$
$\implies W = \left[\left(18 + 9\right) - \left(2 + 3\right)\right]$
$\implies W = 22 J$