Work, Energy and Power

Work done is equal to change in K.E.

$So{W}_1+W_2=\frac{1}{2}M{\left(0.8\sqrt[]{gh}\right)}^2-0$ W₁ ^-> work done by mg

$mgh+W_2=\frac{1}{2}m*0.64gh$  W₂ -> work done by air friction

$W_2=0.32mgh-mgh=-0.68mgh$              

W₂ = -0.68 mgh

Using conservation of linear momentum:

=> 40 * 3m = 60 * m + v * 2m

=> v = 30 m/s

$K{E}_i=\frac{1}{2}*3m*{\left(40\right)}^2$      

$K{E}_f=\frac{1}{2}*m*{\left(60\right)}^2+\frac{1}{2}*2m*{\left(30\right)}^2$     

$\Rightarrow \frac{K{E}_f}{K{E}_i}=\frac{54}{48}$

Fractional change in kinetic energy = 1 $-\frac{K{E}_f}{K{E}_i}=\frac{1}{8}$

$?\frac{P_i=P_f}{m*40}=\frac{m}{2}v\text{'}+\frac{m}{2}*60$

v’ = 20 m/s

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$K{E}_f=\frac{1}{2}\frac{m}{2}\left[v^2+60^2\right]$

$\frac{\Delta KE}{KE}=\frac{\frac{1}{2}m\left(2000\right)-\frac{1}{2}m\left(1600\right)}{\frac{1}{2}m\left(1600\right)}$

$\therefore x=1$

Velocity of bob after collision,

v_{1 $=\sqrt[]{5gR}=\sqrt[]{5*10*0.5}=5m/s$}

From Conservation of Momentum,

$\frac{10}{1000}*v=\frac{-10}{1000}*100+1*5$  

->v = 400 m/s  

W_A = W_B

$\Rightarrow F_A*d*cos45=F_B*d*cos60$     

$\Rightarrow \frac{F_A}{F_B}=\frac{1}{\sqrt[]{2}}=\frac{1}{\sqrt[]{x}}$

=> x = 2

mgh = $ms\Delta T$

$\Rightarrow \Delta T=\frac{gh}{s}=\frac{10*63}{4200}=0.147°C$  

$\left[\lambda *3*g\right]*\frac{3}{2}-\left[\lambda *1*g\right]*\frac{1}{2}=K$         

=> $\lambda \Rightarrow \frac{mass}{length}$

=> $\lambda g\left(\frac{9}{2}-\frac{1}{2}\right)=K$

$\frac{3}{3}*10*4=K$

K = 40 J

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

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

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

= 44%

The two main types of potential energy are:

1. Gravitational Potential Energy - Energy stored due to an object's position in a gravitational field (PE = mgh).
2. Elastic Potential Energy - Energy stored in deformed elastic objects like springs or stretched materials (PE = ½kx²).

The best way to find potential energy is to use the relationship 

${}_{}$ $U_f-U_i=-{\int }_i^f\overrightarrow{F}\cdot d\overrightarrow{r}$

and integrating the conservative force over the path.

For common cases, apply the standard potential energy formulas:

• PE = mgh for gravity
• PE = ½kx² for springs.