Work, Energy and Power

Com (Conservation of Momentum)

Pi = Pf

$\Rightarrow m{u}_1+5m{u}_2=m{v}_1+5m{v}_2$

$\Rightarrow m{u}_1=m{v}_1+5m{v}_2$

${}_{}$ $\Rightarrow u_1+v_1+5v_2$  …….(i)

& $e=1=\frac{v_2-v_1}{U_1-U_2}=\frac{v_2-v_1}{u_1}$ $\frac{{}_{}{}_{}}{{}_{}{}_{}}\frac{{}_{}{}_{}}{{}_{}}$

${}_{}{}_{}{}_{}$  $\Rightarrow u_1=v_2-v_1$ …….(ii)

Equation (i) + (ii)

$k.E_T=\frac{5}{9}{\left(K.E\right)}_i=55.6\mathrm{\%}$

Net force on motor will be

$F_{\text{min}}=[920+68(10\left)\right]g+6000$

$=22000\mathrm{N}$

So, required power for motor

$\begin{array}{rr}{P}_{\text{min}}& ={\stackrel{?}{F}}_{\text{min}}\cdot \stackrel{?}{v}\\ & =22000*3\\ & =66000\text{watt}\end{array}$

$v_i=3*{\left(0\right)}^2+4=4m/s$

$v_F=3*2^2+4=16m/s$

$\omega =\Delta k=\frac{1}{2}m\left[v_F^2-v_i^2\right]$

= $\frac{1}{2}*0.5\left[16^2-4^2\right]=\frac{240}{4}=60J$

$v=\frac{m{v}_0}{M+m}\Rightarrow K_f=\frac{1}{2}\left(M+m\right)v^2=\frac{m{v}_0^2}{2\left(M+m\right)}$

$K_i=\frac{m{v}_0^2}{2}$

Loss in Kinetic energy = $\Delta K=K_i-K_f$

$\Delta K=\frac{1}{2}m{v}_0^2\left(\frac{M}{M+m}\right)=\frac{1}{2}*0.2*100=\left[\frac{9.8}{10}\right]=9.8J$

$\frac{\Delta K}{K_i}*100=\frac{K_f-K_i}{K_i}*100$                

$=\frac{\frac{P_f^2}{2m}-\frac{P_i^2}{2m}}{\frac{P_i^2}{2m}}*100=\frac{P_f^2-P_i^2}{P_i^2}*100$               

$=\frac{{\left(1.2P_i\right)}^2-{\left(P_i\right)}^2}{P_i^2}*100=\frac{1.44P_i^2-P_i^2}{P_i^2}*100$               

 = 44%

$\mathrm{K}\mathrm{E}={\mathrm{P}\mathrm{E}}_1-{\mathrm{P}\mathrm{E}}_2=mg{\mathrm{h}}_1-{\mathrm{m}\mathrm{g}\mathrm{h}}_2$

10

loss in k = Gain in U spring

$\Rightarrow \frac{3E}{4}=\frac{1}{2}k{\left(\frac{25}{100}\right)}^2$

$\Rightarrow \frac{3E}{4}=\frac{k}{32}\Rightarrow k=\frac{3F}{4}*32=24E$                

$\overrightarrow{F}=4x\widehat{i}+3y^2\widehat{j}$

$\begin{array}{l}d\overrightarrow{r}=d\overrightarrow{s}=dx\widehat{i}+dy\widehat{j}\\ {\displaystyle \int dw={\displaystyle \int \overrightarrow{F}.d\overrightarrow{r}}}\\ ={\displaystyle \int \left(4x\widehat{i}+3y^2\widehat{j}\right).\left(dx\widehat{i}+dy\widehat{j}\right)}\\ ={\displaystyle \underset{1}{\overset{2}{\int }}4xdx+{\displaystyle \underset{2}{\overset{3}{\int }}3y^{2}dy}}\\ =\frac{4}{2}{\left[x^2\right]}_1^2+\frac{3}{3}{\left[y^3\right]}_2^3\\ =2[4-1]+[27-8]\end{array}$

= 6 + 19 = 25 J