Figure 7-41shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height$\mathbf{h}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{}\mathbf{m}$, so the cart slides from${\mathbf{x}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{x}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$. During the move, the tension in the cord is a constant$\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. What is the change in the kinetic energy of the cart during the move?

1. The height if the pulley $\mathrm{h}=1.20\mathrm{m}$
2. The displacement of the cart from ${\mathrm{x}}_1=3.00\mathrm{m}\mathrm{to}{\mathrm{x}}_2=1.00\mathrm{m}$
3. The tension in the cord is$25.0\mathrm{N}$

The problem deals with the concept of work done. It includes both forces exerted on the body and the total displacement of the body. The change in the kinetic energy of an object is the work done by the force acting on the object.

Formula:

The change in the kinetic energy of an object is the work done on the object by the force. Here, the cord is pulling the cart, so the tension in the cord does the work on the cart and causes it to displace from x₁ to x₂. During this displacement, the kinetic energy of the cart changes by an amount equal to the work done.

Hence,

$∆\mathrm{KE}=\mathrm{W}$

Now, the work done can be determined as

$\begin{array}{rcl}\mathrm{W}& =& \overrightarrow{\mathrm{F}}∆\mathrm{dx}\\ & =& \overrightarrow{\mathrm{T}}∆\mathrm{dx}\\ & =& \mathrm{Tdx}\mathrm{cos\theta }\end{array}$

Where, angle$\mathrm{\theta }$ is the angle made by the tension with the horizontal.

Using geometry of the figure, we can write

where

$\mathrm{cos\theta }=\frac{\mathrm{x}}{\sqrt{{\mathrm{h}}^2+{\mathrm{x}}^2}}\mathrm{where}\mathrm{x}={\mathrm{x}}_1-{\mathrm{x}}_2$

Now,

$$\begin{gathered} \mathrm{W}=\int \mathrm{T}\mathrm{cos\theta }\mathrm{dx} \\ \mathrm{W}=\mathrm{T}{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\left[\frac{\mathrm{x}}{\sqrt{{\mathrm{h}}^2+{\mathrm{x}}^2}}\right]d\mathrm{x} \end{gathered}$$

Here we can use substitution method to solve this integration.

Let us suppose,${\mathrm{h}}^2+{\mathrm{x}}^2={\mathrm{r}}^2$

So that,

$\begin{array}{rcl}2\mathrm{xdx}+0& =& 2\mathrm{rdr}\\ \mathrm{xdx}& =& \mathrm{rdr}\\ & & \end{array}$

We find the limits as, when$\mathrm{x}={\mathrm{x}}_1$,

$\begin{array}{rcl}\mathrm{r}& =& {\mathrm{r}}_1\\ & =& \sqrt{{\mathrm{h}}^2+{\mathrm{x}}_1^2}\\ & =& \sqrt{{\left(1.20\mathrm{m}\right)}^2+{\left(3.00\mathrm{m}\right)}^2}\\ & =& \sqrt{10.44{\mathrm{m}}^2}\\ & =& 3.23\mathrm{m}\end{array}$

And when, $\mathrm{x}={\mathrm{x}}_2$

$\begin{array}{rcl}\mathrm{r}& =& {\mathrm{r}}_2\\ & =& \sqrt{{\mathrm{h}}^2+{\mathrm{x}}_2^2}\\ & =& \sqrt{{\left(1.20\mathrm{m}\right)}^2+{\left(1.00\mathrm{m}\right)}^2}\\ & =& \sqrt{2.44{\mathrm{m}}^2}\\ & =& 1.56\mathrm{m}\end{array}$

Now we determine the work done as,

$\begin{array}{rcl}\mathrm{W}& =& \mathrm{T}{\int }_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}\left[\frac{\mathrm{x}}{\sqrt{{\mathrm{h}}^2+{\mathrm{x}}^2}}\right]d\mathrm{x}\\ & =& \mathrm{T}{\int }_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\frac{\mathrm{rdr}}{\sqrt{{\mathrm{r}}^2}}\\ \mathrm{W}& =& \mathrm{T}{\int }_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\mathrm{dr}\\ \mathrm{W}& =& \left(25.0\mathrm{N}\right){\left[\mathrm{r}\right]}_{{\mathrm{r}}_1}^{{\mathrm{r}}_2}\\ & =& \left(25.0\mathrm{N}\right)\left(1.56-3.23\right)\\ \mathrm{W}& =& -41.7\mathrm{J}\end{array}$

The change in kinetic energy is $∆\mathrm{KE}=\mathrm{W}=41.7\mathrm{J}$(considering magnitude only).