A 3.0 kg body is at rest on a frictionless horizontal air track when a constant horizontal force $\overrightarrow{\mathbf{F}}$acting in the positive direction of an x axis along the track is applied to the body. A stroboscopic graph of the position of the body as it slides to the right is shown in Fig. 7-25. The force $\overrightarrow{\mathbf{F}}$is applied to the body at t=0, and the graph records the position of the body at 0.50 s intervals. How much work is done on the body by the applied force $\overrightarrow{\mathbf{F}}$between t=0 and t=2.0 s ?

1. Mass, m=3.0 kg.
2. Time interval, t=0 s to t=2.0 s.
3. x₀ is the initial position, which is 0.

Using the formula of the second kinematic equation, we can writetheequation of displacement for any two points. By solving these two equations simultaneously, we can find velocity. Using this velocity, we can computethework done on the body by forcebetween two intervals.

Formula:

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_0+\frac{1}{2}{\mathrm{at}}^2 \\ {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at} \\ \mathrm{W}=∆\mathrm{KE} \end{gathered}$$

The position of a particle can be calculated as,

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_0+{\mathrm{v}}_0\left(\mathrm{t}\right)+\frac{1}{2}{\mathrm{at}}^2$

Here x₀ is the initial position, v₀ is the initial velocity, and a is the acceleration of the particle.

For $\mathrm{x}=0.2\mathrm{m}\mathrm{and}\mathrm{x}=0.8\mathrm{m}$, the above expression can be written as,

$$\begin{gathered} 0.2\mathrm{m}={\mathrm{v}}_0\left(1\mathrm{s}\right)+\frac{1}{2}\mathrm{a}{\left(1\mathrm{s}\right)}^2 \\ 0.8\mathrm{m}={\mathrm{v}}_0\left(2\mathrm{s}\right)+\frac{1}{2}\mathrm{a}{\left(2\mathrm{s}\right)}^2 \end{gathered}$$

From the above two equations, we can calculate,

${\mathrm{v}}_0=0\frac{\mathrm{m}}{\mathrm{s}},\mathrm{a}=0.40\frac{\mathrm{m}}{{\mathrm{s}}^2}$

Now, for the velocity at t=2.0 s

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{i}}+\mathrm{at} \\ {\mathrm{v}}_{\mathrm{f}}=\left(0\frac{\mathrm{m}}{\mathrm{s}}\right)+\left(0.40\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(2.0\mathrm{s}\right) \\ {\mathrm{v}}_{\mathrm{f}}=0.80\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

According to the work-energy theorem, the change in kinetic energy is the work done on the system.

Therefore, work done can be calculated as,

$\begin{array}{rcl}\mathrm{W}& =& ∆\mathrm{KE}\\ & =& {\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}\\ & =& \frac{1}{2}{\mathrm{mv}}_{\mathrm{f}}^2-\frac{1}{2}{\mathrm{mv}}_{\mathrm{i}}^2\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\left(3.0\mathrm{kg}\right){\left(0.80\frac{\mathrm{m}}{\mathrm{s}}\right)}^2-\frac{1}{2}\left(3.0\mathrm{kg}\right){\left(0\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\\ & =& 0.96.\left(1\mathrm{kg}.{\left(\frac{\mathrm{m}}{\mathrm{s}}\right)}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\left({\displaystyle \frac{\mathrm{m}}{\mathrm{s}}}\right)}^2}\right)\\ \mathrm{W}& =& 0.96\mathrm{J}\end{array}$

Therefore, the work done on the body by force data-custom-editor="chemistry" $\overrightarrow{\mathrm{F}}$between the t=0 and t=2.0 s is 0.96 J.