91影视

Given in the question,

The radius of the pully is $R$

The moment of inertia of pully is$I$

The mass of block A is$m_A$

The mass of block B is$m_B$

The coefficient of kinetic friction between the tabletop is${渭}_k$

The law of conservation of energy is, that 鈥渢he initial total energy is always equal to the final total energy of the system鈥�.

$E_F=E_I$

From the conservation of energy, we know

$E_I=E_F$

Initial potential energy +initial kinetic energy= final potential energy + final kinetic energy + work done by friction.

Therefore, we can write

Change in potential energy = Change in kinetic energy +$W_f$

$螖U=螖K+W_f$鈥�(颈)

Since the upper block and the pully does not change their position

Therefore, there is no change in their potential energy.

Potential energy can be given as

$U=mgh$

Where Uis potential energy, m is mass and h is the height.

$螖U=m_{B}gd$

So, the change in potential energy is only due to block B, and since block B is descended by distance d.

Since the system is initially at the rest therefore initial kinetic energy of the system is zero, so the change in kinetic energy is equal to the of the transitional kinetic energy of blocks A, and B and the rotational energy of the pully.

The formula of transitional kinetic energy is

$K_t=\frac{1}{2}m{v}^2$

Where m is mass and v is the linear speed

The formula of rotational kinetic energy

$k_r=\frac{1}{2}I{蝇}^2$

Where I is inertia and $蝇$ is the angular velocity

Since blocks are connected complete system will move with the same linear speed. So, a change in kinetic speed can be given as

$螖K=\frac{1}{2}{m}_A{v}^2+\frac{1}{2}{m}_B{v}^2+\frac{1}{2}I{蝇}^2$

Where v is linear speed and $蝇$is the angular speed of the pully

Since block moves at the distance d

Work done by the friction

$\begin{array}{c}{W}_k=\text{蹿辞谤肠别脳诲}\\ =u_k{m}_{A}gd\end{array}$

Substituting all the values into equation (i)

$\begin{array}{l}螖U=螖K+W_f\\ m_{B}gd=\frac{1}{2}{m}_A{v}^2+\frac{1}{2}{m}_B{v}^2+\frac{1}{2}I{蝇}^2+u_k{m}_{A}gd\end{array}$

We know

$v=r蝇$

Therefore,

$\begin{array}{c}{m}_{B}gd=\frac{1}{2}{m}_A{v}^2+\frac{1}{2}{m}_B{v}^2+\frac{1}{2}I{\left(\frac{v}{R}\right)}^2+u_k{m}_{A}gd\\ v^2\left[\frac{1}{2}{m}_A+\frac{1}{2}{m}_B+\frac{1}{2}\frac{I}{R^2}\right]=m_{B}gd鈭�u_k{m}_{A}gd\\ v^2=\frac{2\left(m_{B}gd鈭�u_k{m}_{A}gd\right)}{\left[m_A+m_B+\frac{I}{R^2}\right]}\\ v=\sqrt{\frac{2gd\left(m_B鈭�u_k{m}_A\right)}{\left[m_A+m_B+\frac{I}{R^2}\right]}}\end{array}$

Hence the speed of block B is$v=\sqrt{\frac{2gd\left(m_B鈭�u_k{m}_A\right)}{\left[m_A+m_B+\frac{I}{R^2}\right]}}$