Q81P - 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. Initially the block is moving downward and the block is moving to the right, both with a speed of . The blocks come to rest after moving 2.00 m. Use the work–energy theorem to calculate the coefficient of kinetic friction between the 8.00-kg block and the table top.

                         

Step-by-Step Solution

Answer

The coefficient of kinetic friction is 0.786.

Step 1: Identification of given data

The given data can be listed below,

• The mass of horizontal block is, ${\mathrm{m}}_{\mathrm{h}}=8.0\mathrm{kg}$
• The mass of vertical block is, ${\mathrm{m}}_{\mathrm{v}}=6.0\mathrm{kg}$
• The speed of the blocks is, $\mathrm{v}=0.900\mathrm{m}/\mathrm{s}$ 
• The distance the block moved is, l = 2.0 m

Step 2: Concept/Significance of kinetic friction.

The ratio of a body's normal force to the kinetic frictional forces between surfaces of contact is known as the coefficient of kinetic friction.

Step 3: Determination of the coefficient of kinetic friction between the 8.00-kg block and the table top.

The free body diagram for the motion of blocks is given below as,

The work done due to the vertical block is given by,

          $$\begin{gathered} W_{v,tot}=W_{gravity}+W_t \\ {m}_{v}gl-Tl=-\frac{1}{2}{m}_v{v}_0^2 \end{gathered}$$

Here, m is the mass of the vertical block, g is the acceleration due to gravity, T is the tension in the string, l is the distance the block moved, and is the velocity of the block.

The work done due to horizontal block is given by,

$$\begin{gathered} W_{h,tot}=W_{tv}+W_f \\ Tl-{\mu }_k{m}_{h}gl=-\frac{1}{2}{m}_h{v}_0^2 \end{gathered}$$

On addition of the equation (i) and (ii) the coefficient of friction is given by,

$$\begin{gathered} m_{v}gl-Tl+Tl-{\mu }_k{m}_{h}gl=-\frac{1}{2}{m}_v{v}_0^2 \\ m_{v}gl-{\mu }_k{m}_{h}gl=-\frac{1}{2}\left(m_v+m_h\right)v_0^2 \\ {\mu }_k{m}_{h}gl=m_v+\frac{1}{2}\left(m_v+m_h\right)v_0^2 \\ {\mu }_k=\frac{m_{v}gl+{\displaystyle \frac{1}{2}}\left(m_v+m_h\right)v_0^2}{m_{h}gl} \end{gathered}$$

Substitute all the values in the above,

$$\begin{gathered} {\mathrm{\mu }}_{\mathrm{k}}=\frac{\left(6\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2.0\mathrm{m}\right)+\left(0.5\right)(6\mathrm{kg}+8\mathrm{kg}){\left(0.900\mathrm{m}/\mathrm{s}\right)}^2}{\left(8\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(2\mathrm{m}\right)} \\ =\frac{\left(117.6+5.67\right)\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}{156.8 \mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =0.786 \end{gathered}$$

Thus, the coefficient of kinetic friction is 0.786.