Answer
The change in the kinetic energy of the cart is $~~41.7~J$
Work Step by Step
Let $\theta$ be the angle that the cord makes above the horizontal.
We can write an expression for the horizontal component of the tension force $F_T$:
$F_x = F_T~cos~\theta$
Note that at each point $x$, the length of the cord between the cart and the pulley is $\sqrt{x^2+(1.2)^2}$
Then, at each point $x$:
$cos~\theta = \frac{x}{\sqrt{x^2+(1.2)^2}} = \frac{x}{\sqrt{x^2+1.44}}$
We can write an expression for the horizontal component of the tension force $F_T$:
$F_x = F_T~cos~\theta$
$F_x = \frac{x~F_T}{\sqrt{x^2+1.44}}$
We can find find the work done on the cart by the cord:
$W = \int^{1.00}_{3.00} \frac{x~F_T}{\sqrt{x^2+1.44}}~dx$
Let $u = x^2+1.44$
$\frac{du}{dx} = 2x$
$dx = \frac{du}{2x}$
When $x = 1.00$, then $u = 2.44$
When $x = 3.00$, then $u = 10.44$
We can find find the magnitude of the work done on the cart by the cord:
$W = \vert \int^{1.00}_{3.00} \frac{x~F_T}{\sqrt{x^2+1.44}}~dx \vert$
$W = \vert \int^{2.44}_{10.44} \frac{x~F_T}{\sqrt{u}}~\frac{du}{2x} \vert$
$W = \vert \int^{2.44}_{10.44} \frac{F_T}{2\sqrt{u}}~du \vert$
$W = \vert F_T~\sqrt{u}~\Big\vert^{2.44}_{10.44} \vert$
$W = \vert (25.0)~(\sqrt{2.44}-\sqrt{10.44}) \vert$
$W = 41.7~J$
The change in the kinetic energy of the cart is $~~41.7~J$.
