Work, Energy and Power

The 1/2 is a result of mathematical calculation, which occurs when we integrate? vdv in the formula of work done according to Newton's second law of motion. Without this, the final result will turn out to be twice of the actual value.

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