(A) Power is defined as $P = Fv$. Since $F = ma$,we have $P = mav$.Using the definition of acceleration $a = v \frac{dv}{dx}$,we can write $P = m v \f

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${\left( {\frac{{3xP}}{m}} \right)^{1/3}}$

(A) Power is defined as $P = Fv$. Since $F = ma$,we have $P = mav$.Using the definition of acceleration $a = v \frac{dv}{dx}$,we can write $P = m v \frac{dv}{dx} v = m v^2 \frac{dv}{dx}$.Rearranging the terms,we get $P dx = m v^2 dv$.Integrating both sides with initial conditions $x=0$ at $v=0$:$\int_0^x P dx = \int_0^v m v^2 dv$.$Px = m \frac{v^3}{3}$.Solving for $v$,we get $v^3 = \frac{3Px}{m}$,which implies $v = {\left( {\frac{{3xP}}{m}} \right)^{1/3}}$.

Class 11 Physics Work, Energy, Power and Collision Power

Difficult

$A$ car of mass $m$ is driven with constant power $P$. If it starts from rest,find its velocity $v$ after traveling a distance $x$.

A
${\left( {\frac{{3xP}}{m}} \right)^{1/3}}$
B
${\left( {\frac{{2xP}}{m}} \right)^{1/2}}$
C
${\left( {\frac{{xP}}{m}} \right)^{1/3}}$
D
${\left( {\frac{{2xP}}{{3m}}} \right)^{1/4}}$

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Work, Energy, Power and Collision

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Water falls on the blades of a turbine at a rate of $100 \ kg/s$ from a height of $100 \ m$. The power generated is . . . . . . $kW$.

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$A$ body is moving along a straight line under the influence of a constant power source. If the relation between the displacement $(s)$ of the body and time $(t)$ is $s \propto t^x$, then $x=$

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$A$ car of mass $m$ is driven with constant power $P$. If it starts from rest,find its velocity $v$ after traveling a distance $x$.

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