Question

A block of mass $m$ attached with an ideal spring of for constant $k$ is placed on a rough inclined plane having inclination $\theta$ with the horizontal and coefficient of friction $\mu =1/2\tan \theta$. Initially the block is held stationary with the spring in its relaxed state, find the maximum extension in the spring if the block is released.

Answer 1

$-\frac{1}{2}k{x}^2+mgx\sin \theta -\mu mgx\cos \theta =0$
$mg\sin \theta -\mu mg\cos \theta =\frac{1}{2}kx\Rightarrow x=\frac{mg\sin \theta }{k}$
Power is defined as the rate of work done
ie., $P=dW/dt...\left(1\right)$
As $dW=\overrightarrow{F}.d\overrightarrow{x}\Rightarrow P=dW/dt=\overrightarrow{F}.\left(d\overrightarrow{x}/dt\right)$
$P=\overrightarrow{F}\overrightarrow{v}...\left(2\right)$
If the force is variable, we calculate the average power as
$P_{av}=\frac{\Delta W}{\Delta t}=\frac{{\int }_0^{t}Pdt}{{\int }_0^{t}dt}$
Power can also be expressed as the rate of change of kinetic energy. Let a body of mass $m$ move with a velocity $v$. the kinetic energy of the body is
$K=\frac{1}{2}m{v}^2$
Now $dK/dt=\frac{1}{2}\frac{d}{dt}m{v}^2=m.\overrightarrow{v}\left(\frac{d\overrightarrow{v}}{dt}\right)=m\left(\frac{d\overrightarrow{v}}{dt}\right)={\overrightarrow{F}}_{net}.\overrightarrow{v}=P$
Therefore, $P=dK/dt$