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 shown above). 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 occurs at the moment when the cord deviates from the vertical by an angle $\theta ={30}^{\circ }$. 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 is $A/100$ JOule. Find the value of A.

Answer 1

Obviously the elongation in the cord, $\Delta l=l_0(\sec \theta -1)$, at the moment the sliding first starts and at the moment horizontal projection of spring force equals the limiting friction.
So, ${\kappa }_1\Delta l\sin \theta =kN$ (1)
(where ${\kappa }_1$ is the elastic constant).
From Newton's law in projection form along vertical direction
${\kappa }_1\Delta l\cos \theta +N=mg$
or, $N=mg-{\kappa }_1\Delta l\cos \theta$ (2)
From (1) and (2),
${\kappa }_1\Delta l\sin \theta =k(mg-{\kappa }_1\Delta l\cos \theta )$
or, ${\kappa }_1=\frac{kmg}{\Delta l\sin \theta +k\Delta l\cos \theta }$
From the equation of the increment of mechanical energy, $\Delta U+\Delta T=A_{fr}$
or, $\left(\frac{1}{2}{\kappa }_1\Delta l^2\right)=A_{fr}$
or, $\frac{kmg\Delta l^2}{2\Delta l(\sin \theta +k\cos \theta )}=A_{fr}$
Thus $A_{fr}=\frac{kmg{l}_0(\sec \theta -1)}{2(\sin \theta -k\cos \theta )}=0.09J$ (on substitution)