Q82P - University Physics with Modern Physics

Question

Consider the system shown in Fig. P6.81. The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the 8.00-kg block and the table top is  ${\mathit{\mu }}_{\mathbf{k}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}$. The blocks are released from rest. Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

                                                       

Step-by-Step Solution

Answer

The velocity of block of mass 6 kg, after it has descended 1.50 m is 2.9 m/s .

Step 1: Identification of given data

The given data can be listed below,

• The mass of the hanging block is, $m_1=6 \mathrm{kg}$ 
• The mass of the resting block is, $m_2=8 \mathrm{kg}$ 
• The coefficient of kinetic friction is, ${\mu }_k=0.250$
• The distance descended by block is, $d=1.50 \mathrm{m}$  

Step 2: Concept/Significance of friction.

The resistance to motion of one item moving in relation to another is known as friction. It results from the electromagnetic attraction of charged particles on two surfaces that are in contact.

Step 3: Determination of the speed of the 6.00-kg block after it has descended 1.50 m

The net work done by the horizontal mass, considering that body is initially at rest, is given as-

$W_{net}=\frac{1}{2}{m}_1{v}^2$                                                                                                                        …(i)

Here, $m_1$ is the mass of horizontal block, v  is the velocity of horizontal block.

The net work done of the block is also given as,

$W_{net}=W_T-W_f$

             $W_T={\mu }_k{m}_{1}gd$                                                               …(ii)

Here, $W_T$  is the work done by tension $\left(T\right)$ in the string.          

                                                                                         

Substitute the given values in the above equation,

$W_T={\mu }_k{m}_{1}gd=\frac{1}{2}{m}_1{v}^2$                                                                                                        …(iii)

The net work done by hanging mass is given by,

$W_{net}=\frac{1}{2}{m}_2{v}^2$                                                                                                                     …(iv)

Here, $m_2$  is the mass of the hanging block, v  is the velocity of the hanging block.

The net work done by the block is also given as,

$$\begin{gathered} W_{net}=W_w-W_T \\ =-W_T+m_{2}gd \end{gathered}$$                                                                                                             …(v)

Comparing equation (iv) & (v),

$-W_T+m_{2}gd=\frac{1}{2}{m}_2{v}^2$                                                                                                        …(vi)

Adding equation (iii) & (vi), we get-

$$\begin{gathered} m_{2}gd-{\mu }_k{m}_{1}gd=\frac{1}{2}\left(m_1+m_2\right)v^2 \\ {v}^2=\frac{2\left(m_{2}gd-{\mu }_k{m}_{1}gd\right)}{\left(m_1+m_2\right)} \\ v=\sqrt{\frac{2gd\left(m_2-{\mu }_k{m}_1\right)}{\left(m_1+m_2\right)}} \end{gathered}$$  

Substitute the given values in the above,

$$\begin{gathered} v=\sqrt{\frac{2\left(9.8 m/s^2\right)\left(1.50 m\right)\left(6 kg-\left(0.25\times 8 kg\right)\right)}{\left(6+8\right)}} \\ =2.9 m/s \end{gathered}$$  

Thus, the velocity of block of mass 6 kg is 2.9 m/s .