91Ó°ÊÓ

The given data can be listed below as:

• The value of mass$m_1$ is$m_1=20.0\text{ k²µ}$ .
• The angle of inclination of the block is $\mathrm{Î\pm }={53.1}^{∘}$.
• The coefficient of the kinetic friction is ${\mathrm{μ}}_k=0.40$.
• The height descended by the second mass is .$h=12.0\text{″¾}$
• The time taken for the mass to descend is $t=3.00\text{ s}$.

The mass is described as the quantity of the matter inside a particular object. The mass of an object is directly proportional to the force exerted by that object.

The free body diagram of the system is drawn below:

The equation of the acceleration of the system is expressed as:

$h=ut+\frac{1}{2}a{t}^2$

Here,$h$is the height descended by the second mass,$u$is the initial speed of the block,$t$is the time taken for the mass to descend and $a$is the acceleration of the block.

As initially the block was at rest, then the initial velocity of the block is zero.

Substitute$0$for $u$, $3.00\text{ s}$for and$12.0\text{″¾}$for $h$in the above equation.

$\begin{array}{c}12.0\text{″¾}=\left(0\right)t+\frac{1}{2}a{\left(3.00\text{ s}\right)}^2\\ 12.0\text{″¾}=\frac{1}{2}a\left(9.00{\text{ s}}^2\right)\\ 12.0\text{″¾}=a\left(4.5{\text{ s}}^2\right)\\ a=2.6{\text{″¾/s}}^{\text{2}}\end{array}$

According to the free body diagram, the equation of the net force is expressed as:

$T−f−m_{1}g\sin \mathrm{Î\pm }=m_{1}a$ …(1)

Here,$T$is the tension of the block,$f$is the frictional force,$m_1$is the mass of the first block,$g$is the acceleration due to gravity and$\mathrm{Î\pm }$is the angle of inclination of the block.

The equation of the frictional force is expressed as:

$f={\mathrm{μ}}_k{m}_{1}g\cos \mathrm{Î\pm }$

Here,${\mathrm{μ}}_k$is the coefficient of the kinetic friction.

Substitute ${\mathrm{μ}}_k{m}_{1}g\cos \mathrm{Î\pm }$for$f$in the above equation.

$\begin{array}{c}T−{\mathrm{μ}}_k{m}_{1}g\cos \mathrm{Î\pm }−m_{1}g\sin \mathrm{Î\pm }=m_{1}a\\ T={\mathrm{μ}}_k{m}_{1}g\cos \mathrm{Î\pm }+m_{1}g\sin \mathrm{Î\pm }+m_{1}a\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}T=\left(0.40\right)\left(20.0\text{ k²µ}\right)\left(9.8{\text{″¾/s}}^{\text{2}}\right)\cos {53.1}^{∘}+\left(20.0\text{ k²µ}\right)\left(9.8{\text{″¾/s}}^{\text{2}}\right)\sin {53.1}^{∘}+\left(20.0\text{ k²µ}\right)\left(2.6{\text{″¾/s}}^{\text{2}}\right)\\ =\left(78.4{\text{ k²µ.m/s}}^{\text{2}}\right)\left(0.6004\right)+\left(196{\text{ k²µ.m/s}}^{\text{2}}\right)\left(0.799\right)+\left(52{\text{ k²µ.m/s}}^{\text{2}}\right)\\ =47.07{\text{ k²µ.m/s}}^{\text{2}}+156.604{\text{ k²µ.m/s}}^{\text{2}}+52{\text{ k²µ.m/s}}^{\text{2}}\\ =255.674{\text{ k²µ.m/s}}^{\text{2}}\end{array}$

The equation of the mass $m_2$is expressed as:

$\begin{array}{c}T−m_{2}g=−m_{2}a\\ m_2(g−a)=T\\ m_2=\frac{T}{g−a}\end{array}$

Here,$m_2$ is the mass of the second particle.

Substitute$255.674{\text{ k²µ.m/s}}^{\text{2}}$ for$T$ ,$9.8{\text{″¾/s}}^{\text{2}}$ for $g$and $2.6{\text{″¾/s}}^{\text{2}}$for$a$ in the above equation.

$\begin{array}{c}{m}_2=\frac{255.674{\text{ k²µ.m/s}}^{\text{2}}}{9.8{\text{″¾/s}}^{\text{2}}−2.6{\text{″¾/s}}^{\text{2}}}\\ =\frac{255.674{\text{ k²µ.m/s}}^{\text{2}}}{7.2{\text{″¾/s}}^{\text{2}}}\\ =35.503\text{ k²µ}\end{array}$

Thus, the mass$m_2$ must be of $35.503\text{ k²µ}$.