91Ó°ÊÓ

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,

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

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