Q5-74P - University Physics with Modern Physics

Question

In Fig. P5.74, and $m_1=20.0\text{\hspace{0.17em}kg}$. The coefficient of kinetic friction between the block of mass $m_1$and the incline is . What must be the mass $m_2$ of the hanging block if it is to descend $12.0\text{\hspace{0.17em}m}$in the first $3.00\text{\hspace{0.17em}s}$after the system is released from rest?

Step-by-Step Solution

Answer

The mass $m_2$ must be of  $35.503\text{\hspace{0.17em}kg}$.

Step 1: Identification of given data

The given data can be listed below as:

• The value of mass $m_1$ is $m_1=20.0\text{\hspace{0.17em}kg}$ .
• The angle of inclination of the block is $\alpha ={53.1}^{\circ }$.
• The coefficient of the kinetic friction is ${\mu }_k=0.40$.
• The height descended by the second mass is .$h=12.0\text{\hspace{0.17em}m}$
• The time taken for the mass to descend is $t=3.00\text{\hspace{0.17em}s}$.

Step 2: Significance of the mass

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.

Step 3: Determination of the mass m 2

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 $u$, $3.00\text{\hspace{0.17em}s}$ for and $12.0\text{\hspace{0.17em}m}$ for $h$ in the above equation.

$\begin{array}{c}12.0\text{\hspace{0.17em}m}=\left(0\right)t+\frac{1}{2}a{\left(3.00\text{\hspace{0.17em}s}\right)}^2\\ 12.0\text{\hspace{0.17em}m}=\frac{1}{2}a\left(9.00{\text{\hspace{0.17em}s}}^2\right)\\ 12.0\text{\hspace{0.17em}m}=a\left(4.5{\text{\hspace{0.17em}s}}^2\right)\\ a=2.6{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$

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

             $T-f-m_{1}g\sin \alpha =m_{1}a$                                                                            …(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 $\alpha$ is the angle of inclination of the block.

 

The equation of the frictional force is expressed as:

 

$f={\mu }_k{m}_{1}g\cos \alpha$

 

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

 

Substitute ${\mu }_k{m}_{1}g\cos \alpha$ for $f$ in the above equation.

 $\begin{array}{c}T-{\mu }_k{m}_{1}g\cos \alpha -m_{1}g\sin \alpha =m_{1}a\\ T={\mu }_k{m}_{1}g\cos \alpha +m_{1}g\sin \alpha +m_{1}a\end{array}$

Substitute the values in the above equation.

 $\begin{array}{c}T=\left(0.40\right)\left(20.0\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\cos {53.1}^{\circ }+\left(20.0\text{\hspace{0.17em}kg}\right)\left(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\sin {53.1}^{\circ }+\left(20.0\text{\hspace{0.17em}kg}\right)\left(2.6{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)\\ =\left(78.4{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)\left(0.6004\right)+\left(196{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)\left(0.799\right)+\left(52{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\right)\\ =47.07{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}+156.604{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}+52{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\\ =255.674{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}\end{array}$

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 $255.674{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}$ for $T$ ,$9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for $g$ and $2.6{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for $a$ in the above equation.

 

$\begin{array}{c}{m}_2=\frac{255.674{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}}{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}-2.6{\text{\hspace{0.17em}m/s}}^{\text{2}}}\\ =\frac{255.674{\text{\hspace{0.17em}kg.m/s}}^{\text{2}}}{7.2{\text{\hspace{0.17em}m/s}}^{\text{2}}}\\ =35.503\text{\hspace{0.17em}kg}\end{array}$

 

Thus, the mass $m_2$ must be of $35.503\text{\hspace{0.17em}kg}$.