Work Power & Energy

We know that $F \times v = $ Power

According to the question, power is constant.

$\therefore$ $F \times v = c\,\,\,\,$ where $c=$ constant

$\therefore$ $m{{dv} \over {dt}} \times v = c$ $\,\,\,\,\left( \, \right.$ $\therefore$ $\left. {F = ma = {{mdv} \over {dt}}\,\,} \right)$

$\therefore$ $m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$ $\therefore$ ${1 \over 2}m{v^2} = ct$

$\therefore$ $v = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$

${{dx} \over {dt}} = \sqrt {{{2c} \over m}} \times {t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$ where $v = {{dx} \over {dt}}$

$\therefore$ $\int\limits_0^x {dx = \sqrt {{{2c} \over m}} } \times \int\limits_0^t {{t^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$
$x = \sqrt {{{2c} \over m}} \times {{2{t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{\raise0.5ex\hbox{$\scriptstyle 3$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}$

The elastic potential energy

$ = {1 \over 2} \times $ Force $ \times $ extension

$= {1 \over 2} \times 200 \times 0.001 = 0.1\,J$

When we extend the spring by $dx$ then the work done

$dW = k\,x\,dx$

Applying integration both sides we get,

$\therefore$ $W = k\int\limits_{0.05}^{0.15} {x\,dx} $

$ = {{800} \over 2}\left[ {{{\left( {0.15} \right)}^2} - {{\left( {0.05} \right)}^2}} \right] $

$= 8\,J$

Assume the ball of mass m is projected with a speed u. Then the kinetic energy(E) at the point of projection = ${1 \over 2}m{u^2}$

At highest point of flight only horizontal component of velocity $u\cos \theta $ present as at highest point vertical component of velocity is = 0.

Note : The horizontal component of velocity does not change in entire projectile motion.

At highest point the velocity is = $u\cos \theta $ = $u\cos 45^\circ $ = ${u \over {\sqrt 2 }}$

$\therefore$ The kinetic energy at the height point = ${1 \over 2}m{\left( {{u \over {\sqrt 2 }}} \right)^2}$

= ${1 \over 2}m{u^2} \times {1 \over 2}$ = ${E \over 2}$

We know the work energy theorem, $W = \Delta K = FS$

For first penetration, by applying work energy theorem we get,

${1 \over 2}m{v^2} - {1 \over 2}m{\left( {{v \over 2}} \right)^2} = F \times 3\,\,...(i)$

For second penetration, by applying work energy theorem we get,

${1 \over 2}m{\left( {{v \over 2}} \right)^2} - 0 = F \times S\,...(ii)$

On dividing $(ii)$ by $(i)$

${{1/4} \over {3/4}} = S/3$

$\therefore$ $S = 1\,cm$