TS EAMCET 5-Aug-2021 Shift 1 Question Paper

Given, constant applied force, $F=5\,\text{N}$, Mass of particle, $m=500\,\text{g}=\frac{500}{1000}=0.5\,\text{kg}$ Displacement, $s=5\,\text{m}$ Initial speed, $u=0\,\text{ms}^{-1}$ Final speed $=v$ As we know that, $\text{Power}(P)=\frac{\text{Work}(W)}{\text{Time}(t)}$. . . (i) According to work-energy theorem, Work done, $W=\frac{1}{2}m(v^{2}-u^{2})$ . . . (ii) and $F=ma$ where, $a$ is acceleration. $\therefore\quad a=\frac{F}{m}$ $=\frac{5}{0.5}=\frac{50}{5}=10\,\text{ms}^{-2}$ and by using third equation of motion $\Rightarrow\quad v^{2}-u^{2}=2as$ $\Rightarrow\quad v^{2}=2as$ $v=\sqrt{2as}$ $\therefore v=\sqrt{2\times10\times5}=\sqrt{100}$ $=10\,\text{ms}^{-1}$ Substituting the values in Eq. (ii), we get $W=\frac{1}{2}\times0.5(10^{2}-0^{2})$ $=\frac{1}{4}(100)=25\,\text{J}$ . . . (iii) According to first equation of motion, $v=u+at$ $\Rightarrow\quad t=\frac{v-u}{a}$ $\Rightarrow\quad t=\frac{10-0}{10}=1\,\text{s}$ . . . (iv) Now, from Eqs. (i), (iii) and (iv), we get $P=\frac{W}{t}$ $=\frac{25}{1}=25\,\text{W}$