Work, Energy and Power

Potential energy is best defined as a stored energy in a system. This energy is stored due to the configuration or position of the objects within a conservative force field. Potential energy in physics tells us that it's the capacity of the work done when the system is allowed to move from that position to the reference point.  

Com

m1v + m2 * 0 = m1v1 + m2v2

75v = $45\frac{v}{3}+50*10$ $\frac{}{}$

$\Rightarrow 75*\frac{2v}{3}=50*10$

$\Rightarrow v=\frac{3*50*10}{75*2}=10m/sec$

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

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

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