Question

In the system shown in figure above, the masses of the bodies are known to be $m_1$ and $m_2$, the coefficient of friction between the body $m_1$ and the horizontal plane is equal to $k$, and a pulley of mass $m$ is assumed to be a uniform disc. The thread does not slip over the pulley. At the moment $t=0$ the body $m_2$ starts descending. Assuming the mass of the thread and the friction in the axle of the pulley to be negligible, find the work performed by the friction forces acting on the body $m_1$ over the first $t$ seconds after the beginning of motion.

• A. $A=-\frac{k{m}_1(m_1-k{m}_1)g^2{t}^2}{m+2(m_1+m_2)}$
• B. $A=-\frac{k{m}_1(m_1+k{m}_1)g^2{t}^2}{m+2(m_1-m_2)}$
• C. $A=\frac{k{m}_1(m_1+k{m}_1)g^2{t}^2}{m+2(m_1+m_2)}$
• D. None of these

Answer 1

Answer: (A)

$A=-\frac{k{m}_1(m_1-k{m}_1)g^2{t}^2}{m+2(m_1+m_2)}$

As the system ($m+m_1+m_2$) is under constant forces, the acceleration of body $m_1$ and $m_2$ is constant. In addition to it the velocities and accelerations of bodies $m_1$ and $m_2$ are equal in magnitude (say $v$ and $w$) because the length of the thread is constant.
From the equation of increment of mechanical energy i.e.
$\Delta T+\Delta U=A_{fr}$, at time $t$, when block $m_1$ is distance $h$ below from initial position corresponding to $t=0$, $\frac{1}{2}(m_1+m_2)v^2+\frac{1}{2}\left(\frac{m{R}^2}{2}\right)\frac{v^2}{R^2}-m_{2}gh=-k{m}_{1}gh$ (1)
(an angular velocity $\omega =\frac{v}{R}$ for no slipping of thread.)
But $v^2=2wh$
So using it in (1), we get
$w=\frac{2(m_2-k{m}_1)g}{m+2(m_1+m_2)}$ (2)
Thus the work done by the friction force on $m_1$
$A_{fr}=-k{m}_{1}gh=-k{m}_{1}g\left(\frac{1}{2}w{t}^2\right)$
$=-\frac{k{m}_1(m_1-k{m}_1)g^2{t}^2}{m+2(m_1+m_2)}$