Question 218: Work, Energy and Power & Circular Motion

Question

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time $'t'$ is proportional to

Accepted Answer

We know that $F 	imes v = $ Power
According to the question, power is constant.
$	herefore$ $F 	imes v = c\,\,\,\,$ where $c=$ constant
$	herefore$ $m{{dv} \over {dt}} 	imes v = c$ $\,\,\,\,\left( \, ight.$ $	herefore$ $\left. {F = ma = {{mdv} \over {dt}}\,\,} ight)$
$	herefore$ $m\int\limits_0^v {vdv = c\int\limits_0^t {dt} } \,\,\,\,\,\,\,\,\,$ $	herefore$ ${1 \over 2}m{v^2} = ct$
$	herefore$ $v = \sqrt {{{2c} \over m}}  	imes {t^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$
${{dx} \over {dt}} = \sqrt {{{2c} \over m}}  	imes {t^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}\,\,\,\,$ where $v = {{dx} \over {dt}}$
$	herefore$ $\int\limits_0^x {dx = \sqrt {{{2c} \over m}} }  	imes \int\limits_0^t {{t^{{aise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} dt$
$x = \sqrt {{{2c} \over m}}  	imes {{2{t^{{aise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over 3} \Rightarrow x \propto {t^{{aise0.5ex\hbox{$\scriptstyle 3$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}$

Answer

${t^{3/4}}$

Answer

${t^{3/2}}$

Answer

${t^{1/4}}$

Answer

${t^{1/2}}$

Work, Energy and Power & Circular Motion

A body is moved along a straight line by a machine delivering a constant power. The distance moved by the body in time 't' is proportional to

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