91Ó°ÊÓ

The given data can be listed below as:

The mass of the first block is${\mathrm{m}}_1$.

The mass of the first block is${\mathrm{m}}_2$.

The value of the slope angle is $\mathrm{Î\pm }$.

The static friction’s coefficient is ${\mathrm{μ}}_{\mathrm{s}}$.

The kinetic friction’s coefficient is ${\mathrm{μ}}_{\mathrm{k}}$.

The tension is a force required to pull a string to transmit the force in the axial direction. Tension is described as the pair of action and reaction forces.

The free body diagram of the block moving upward is drawn below:

The downward tension of the cord’s is${\mathrm{m}}_2\mathrm{g}$where${\mathrm{m}}_2$is the mass of the second block and g is the acceleration due to gravity. However, the tension should overcome the friction as well as the gravitational force’s component in order to move the block at a constant speed. As the mass${\mathrm{m}}_1$is moving at an inclined direction then the weight of the block has two components such as${\mathrm{m}}_1\mathrm{²\mu ³¦´Ç²õÎ\pm }$and${\mathrm{m}}_{1}g\sin \mathrm{Î\pm }$respectively. The frictional force is ${\mathrm{μ}}_k{m}_1\cos \mathrm{Î\pm }$.

The summation of the forces in the x direction is zero.

The equation of the summation of the forces in the direction is expressed as:

$\underset{}{∑}{\mathrm{F}}_{\mathrm{x}}=0$

${\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_1\mathrm{²\mu ²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}{\mathrm{m}}_1\mathrm{²\mu ³¦´Ç²õÎ\pm }$

Here,$\underset{}{∑}{\mathrm{F}}_{\mathrm{x}}$is the summation of the forces in the x direction,${\mathrm{m}}_2$is the mass of the second block,${\mathrm{m}}_1$is the mass of the first block,$\mathrm{Î\pm }$is the angle between the first and the second masses and g is the acceleration due to gravity.

The above equation can be expressed as:

$$\begin{gathered} {\mathrm{m}}_2\sin ={\mathrm{m}}_1\mathrm{²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ³¾}}_1\mathrm{³¦´Ç²õÎ\pm } \\ ={\mathrm{m}}_1\left(\mathrm{²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm }\right) \end{gathered}$$

Thus, the value of the${\mathrm{m}}_2$ when the block moves up the plane is${\mathrm{m}}_1\left(\mathrm{²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm }\right)$

The free body diagram of the block moving downward has been drawn below:

The downward tension of the cord’s is ${\mathrm{m}}_2\mathrm{g}$where ${\mathrm{m}}_2$is the mass of the second block and g is the acceleration due to gravity. However, the tension should overcome the friction as well as the gravitational force’s component in order to move the block at a constant speed. As the mass is moving at an inclined direction then the weight of the block has two components such as and respectively. The frictional force is ${\mathrm{μ}}_k{m}_{1}g\cos \mathrm{Î\pm }$.

The summation of the forces in the direction is zero.

The equation of the summation of the forces in the direction is expressed as:

$$\begin{gathered} \underset{}{∑}{\mathrm{F}}_{\mathrm{x}}=0 \\ {\mathrm{m}}_2\mathrm{g}={\mathrm{m}}_1\mathrm{²\mu ²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}{\mathrm{m}}_1\mathrm{²\mu ³¦´Ç²õÎ\pm } \end{gathered}$$

Here,$\underset{}{∑}{\mathrm{F}}_{\mathrm{x}}$is the summation of the forces in the x direction,${\mathrm{m}}_2$is the mass of the second block,${\mathrm{m}}_1$is the mass of the first block,$\mathrm{Î\pm }$is the angle between the first and the second masses and g is the acceleration due to gravity.

The above equation can be expressed as:

$$\begin{gathered} {\mathrm{m}}_2={\mathrm{m}}_1\mathrm{²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm } \\ ={\mathrm{m}}_1\left(\mathrm{²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm }\right) \end{gathered}$$

Thus, the value of the${\mathrm{m}}_2$ when the block moves down the plane is ${\mathrm{m}}_1\left(\mathrm{²õ¾\pm ²ÔÎ\pm }+{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm }\right)$.

According to the above answers, when the block moves upward, it has the largest value. Moreover, when the block moves downward, then it has lowest value. Hence, the range of should lie between these two values.

Thus, the range of values${\mathrm{m}}_2\mathrm{is}{\mathrm{m}}_1\left(\mathrm{²õ¾\pm ²ÔÎ\pm }-{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm }\right)<{\mathrm{m}}_2<{\mathrm{m}}_1\left(\mathrm{²õ¾\pm ²ÔÎ\pm }-{\mathrm{μ}}_{\mathrm{k}}\mathrm{³¦´Ç²õÎ\pm }\right)$