A particle of mass m is driven by a machine t | vedclass

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(A) The power $P$ delivered to the particle is constant,so $P = k$.Since $P = \frac{dW}{dt}$,we have $dW = k dt$.Integrating both sides from $0$ to $t

Vedclass · Accepted Answer

$\sqrt {\frac{{mk}}{2}} \;{t^{ - \frac{1}{2}}}$

Vedclass · Answer

(A) The power $P$ delivered to the particle is constant,so $P = k$.Since $P = \frac{dW}{dt}$,we have $dW = k dt$.Integrating both sides from $0$ to $t$,we get $W = kt$.According to the work-energy theorem,the work done is equal to the change in kinetic energy:$W = \Delta K = \frac{1}{2}mv^2 - 0$.Equating the two expressions for $W$: $kt = \frac{1}{2}mv^2$,which gives $v = \sqrt{\frac{2kt}{m}}$.The acceleration $a$ is the derivative of velocity with respect to time:$a = \frac{dv}{dt} = \frac{d}{dt} \left( \sqrt{\frac{2k}{m}} t^{1/2} \right) = \sqrt{\frac{2k}{m}} \cdot \frac{1}{2} t^{-1/2} = \sqrt{\frac{k}{2mt}}$.The force $F$ acting on the particle is $F = ma$:$F = m \cdot \sqrt{\frac{k}{2mt}} = \sqrt{\frac{m^2 k}{2mt}} = \sqrt{\frac{mk}{2t}} = \sqrt{\frac{mk}{2}} t^{-1/2}$.

$A$ particle of mass $m$ is driven by a machine that delivers a constant power $k$ watts. If the particle starts from rest,the force on the particle at time $t$ is:

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