Question

A horizontal plane supports a plank with a bar of mass $m=1.0kg$ placed on it and attached by a light elastic non-deformed cord of length $l_0=40cm$ to a point O (figure). The coefficient of friction between the bar and the plank equals $k=0.20$. The plank is slowly shifted to the right until the bar starts sliding over it, it occur at the moment when the cord deviates from the vertical by an angle $\theta ={30}^o$. Find the work that has been performed by that moment by the friction force acting on the bar in the reference frame fixed to the plane.

Answer 1

$N+T\cos \theta =mg...\left(i\right)$
$F_f-T\sin \theta =\mathrm{0...}\left(ii\right)$
$\therefore F_f=T\sin \theta ....\left(iii\right)$
$N=mg-T\cos \theta ....\left(iv\right)$
Further
$T=K(l-l_0)=k{l}_0(\sec \theta -1)$
$x=l_0\tan \theta$
$dx=l_0{\sec }^2\theta d\theta$
Work done $dW={\overrightarrow{F}}_{f}d\overrightarrow{s}=T\sin \theta \left(dx\right)$
$W=\int T\sin \theta \left(dx\right)$
$W=k{l}_0^2(\tan \theta -\sin \theta ){\sec }^2\theta d\theta =k{l}_0^2{[\frac{{\left(\tan \theta \right)}^2}{2}-\frac{1}{\cos \theta .}]}_0^{{30}^o}=k{l}_0^2\times 0.012$
when $\theta ={30}^o,k{l}_0(\sec {30}^o-1)=\frac{0.2mg}{\frac{1}{2}+0.2\times \frac{\sqrt{3}}{2}}\Rightarrow k{l}_0=1.92mg$
$\therefore W=\left(0.023\right)mg{l}_0=0.092J$