91影视

Given in the question,

The radius of the pully, $r=0.160\text{鈥夆赌塵}\mathrm{鈩�}$

The moment of inertia of the pully, $I=0.380{\text{鈥夆赌塳g.m}}^{\text{2}}$

The mass of the block first block$m_1=4.00\text{鈥夆赌塳g}$

The mass of the second block$m_2=2.00\text{鈥夆赌塳g}$

Height of the block from the floor $h_1=5\text{鈥�}m$

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.

$U_i+K{E}_i=U_f+K{E}_f$鈥�(颈)

Initial potential energy will be the sum of the initial potential energy of the pully and the two blocks

The formula of potential energy is,

$U=mgh$

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

Since the height of the pully and rope from the floor do not change,

So, their potential energy will be the same therefore we can take their initial and final potential energy to zero

Therefore, the initial potential energy of the system can be given as

The initial potential energy,

$\begin{array}{l}{U}_i=U_{i\left(Pully\right)}+U_{i\left(rope\right)}+m_1{h}_{1}g+m_2{h}_{2}g\\ U_i=0+0+\left(4.00\text{鈥夆赌塳g}\right)\left(5\text{鈥�}m\right)\left(9.8\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)+\left(2.00\text{鈥夆赌塳g}\right)\left(0\text{鈥�}m\right)\left(9.8\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)\\ U_i=196\text{鈥夆赌塉}\end{array}$

Initial kinetic energy will be the sum of the initial kinetic energy of the rope and pully and the energy of the two-block

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

Initially, the system was at rest therefore initial kinetic energy.

$K{E}_i=0$

Finally, the first book is on the floor therefore its height is zero, and the height of the second block is 5 m.

Final potential energy

$\begin{array}{l}{U}_f=U_{f\left(Pully\right)}+U_{f\left(rope\right)}+m_1{h}_{1}g+m_2{h}_{2}g\\ U_f=0+0+\left(4.00\text{鈥夆赌塳g}\right)\left(0m\right)\left(9.8\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)+\left(2.00\text{鈥夆赌塳g}\right)\left(5\text{鈥�}m\right)\left(9.8\text{鈥夆赌塵}/{\text{s}}^{\text{2}}\right)\\ U_f=98\text{鈥夆赌塉}\end{array}$

Final kinetic energy will be the sum of the transitional kinetic energy of the two-block and rotational kinetic energy of the pully since the mass of the rope is negligible, we can ignore its kinetic energy

Final kinetic energy

$\begin{array}{c}K{E}_f=K_1+K_2+K_{r\left(pully\right)}\\ =\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{2}I{蝇}^2\\ =\frac{1}{2}\left(4.00\text{鈥夆赌塳g}\right)v^2+\frac{1}{2}\left(2.00\text{鈥夆赌塳g}\right)v^2+\frac{1}{2}\left(0.380{\text{鈥夆赌塳g.m}}^{\text{2}}\right){蝇}^2\end{array}$

Since all the blocks and pully are connected by the rope, therefore, they will move with the same linear speed

Therefore, we can use the formula $蝇=\frac{v}{r}$

$\begin{array}{l}K{E}_f=\frac{1}{2}\left(4.00\text{鈥夆赌塳g}\right)v^2+\frac{1}{2}\left(2.00\text{鈥夆赌塳g}\right)v^2+\frac{1}{2}\left(0.380{\text{鈥夆赌塳g.m}}^{\text{2}}\right){\left(\frac{v}{r}\right)}^2\\ K{E}_f=\frac{1}{2}\left(4.00\text{鈥夆赌塳g}\right)v^2+\frac{1}{2}\left(2.00\text{鈥夆赌塳g}\right)v^2+\frac{1}{2}\left(0.380{\text{鈥夆赌塳g.m}}^{\text{2}}\right){\left(\frac{v}{0.160\text{鈥夆赌塵}}\right)}^2\\ K{E}_f=\left[\frac{1}{2}\left(4.00\text{鈥夆赌塳g}\right)+\frac{1}{2}\left(2.00\text{鈥夆赌塳g}\right)+\frac{1}{2}\frac{\left(0.380{\text{鈥夆赌塳g.m}}^{\text{2}}\right)}{0.160\text{鈥夆赌塵}}\right]v^2\\ K{E}_f=10.42\text{鈥夆赌�}{v}^2\text{鈥夆赌塉}\end{array}$

Now substituting all the values of energy into the equation (i)

$\begin{array}{c}{U}_i+K{E}_i=U_f+K{E}_f\\ 196\text{鈥夆赌塉}+0=98\text{鈥夆赌塉+鈥�}10.42\text{鈥夆赌�}{v}^2\text{鈥夆赌塉}\\ v\text{=3.067鈥夆赌塵}/\text{s}\end{array}$

Hence the velocity of the block just before it strikes the floor is 3.067m/s