Work, Energy and Power

W = mgh

$=80*9.8*\left(\frac{80}{100}\right)$

$=627.2J$

$\begin{array}{l}\Rightarrow w_f+w_g=0\\ \Rightarrow w_f=-627.2J\end{array}$  

Let ball starts its motion with horizontal velocity v₀, so with the help of conservation of mechanical energy, we can write

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{x}^2\Rightarrow v_0=x\sqrt[]{\frac{k}{m}}=0.05*\sqrt[]{\frac{100}{0.1}}=0.5*\sqrt[]{10}m/s$              

t = Time required to fall the ball =    $\sqrt[]{\frac{2h}{g}}=2*\sqrt[]{\frac{1}{10}}s$

^{$=d=v_0*t=0.5*\sqrt[]{10}*2*\sqrt[]{\frac{1}{10}}=1m$}              

According to question, we can write

$p=\sqrt[]{2mK}\Rightarrow \frac{p_2}{p_1}=\sqrt[]{\frac{K_2}{K_1}}\Rightarrow \frac{p_2}{p}=\sqrt[]{\frac{4K_1}{K_1}}=2\Rightarrow p_2=2p$

The percentage change in its momentum = $\frac{\Delta p}{p}*100=\frac{p}{p}*100=100\mathrm{\%}$

Maximum energy = 10 J

$\frac{1}{2}K{x}^2=10$              

K = 5

Given T_pendulum = T_spring

$2\pi \sqrt[]{\frac{\mathcal{l}}{g}}=2\pi \sqrt[]{\frac{m}{K}}$             

$\sqrt[]{\frac{4}{g}}=\sqrt[]{\frac{5}{5}}$           

g = 4m/s²

Energy required to melt

Q = $MS\Delta T+ML$  

$\Rightarrow 10^{-1}*2*10^3*10+10^{-1}*3.33*10^5$      

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

  $Q=3.53*10^4={\left(\frac{1}{2}\right)}^2*\left(4*10^3\right)*t$

$t=\frac{3.53*10^4*4}{4*10^3}=35.3sec$             

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$=0-\frac{1}{2}m{V}^2$              

As 90% of K.E. is losed by friction so that

$-\frac{90}{100}\left(\frac{1}{2}m{V}^2\right)-\frac{1}{2}K{x}^2=-\frac{1}{2}m{V}^2$                

-K -> -16 * 10⁵

K = 16 * 10⁵

Object is moving in upward direction with constant velocity so in upward motion (+2N) and for downward motion (-2N) So option (1) is correct representation.

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 $-\frac{K{E}_f}{K{E}_i}=\frac{1}{8}$