Q96P - University Physics with Modern Physics

Question

Two blocks, with masses 4.00 kg and 8.00 kg, are connected by a string and slide down a 30.0° inclined plane (Fig. P5.96). The coefficient of kinetic friction between the 4.00-kg block and the plane is 0.25; that between the 8.00-kg block and the plane is 0.35. Calculate 

(a) the acceleration of each block and 

(b) the tension in the string. 

 

(c) What happens if the positions of the blocks are reversed, so that the 4.00-kg block is uphill from the 8.00-kg block?

Step-by-Step Solution

Answer

(a) The acceleration of each block is $2.212m/s^2$ .

(b) The tension in the string is $2.265 N$ .

(c) The acceleration of the blocks is $2.78m/s^2$ , and $1.93m/s^2$ , when the blocks are reversed.

Step 1: Identification of given data

The given data is as follows:

 

• The mass of the first block is $m_1=4 kg$ .
• The mass of the second block is $m_2=8 kg$ .
• The angle is $\theta =30°$ .
• The coefficient of kinetic friction between the first block and the plane is ${\mu }_{k_1}=0.25$ .
• The coefficient of kinetic friction between the second block and the plane is ${\mu }_{k_2}=0.35$ .

Step 2: Concept/Significance Kinetic friction force

The kinetic force occurs when the object is in motion. The expression of the kinetic friction force is given by,

${\mathit{f}}_{\mathbf{k}}\mathbf{=}{\mathit{\mu }}_{\mathbf{k}}\mathit{N}$  

Here,  ${\mathit{\mu }}_{\mathbf{k}}$ is the coefficient of kinetic friction, and  N is a normal force.

Step 3: Find the acceleration of each block (a)

The forces acting on mass $m_1$  are shown in the following figure.

From the above figure, the relation of forces on the horizontal direction is given by,

 

$m_{1}a=m_{1}g \sin \theta -T-f_{k_1}$  

 

Here,  $m_1$ is the mass of the first block,  a is acceleration,  g is the acceleration due to gravity, $\theta$  is the angle,  T is the tension force, and  $f_{k_1}$ is the kinetic friction due to the first block.

 

The kinetic friction is given by,

 

$f_{k_1}={\mu }_{k_1}{N}_1$  

 

Here ${\mu }_{k_1}$  is the coefficient of kinetic friction and  $N_1$ is the normal force.

 

Substitute  $m_{1}g \cos \theta$ for $N_1$  in the equation $f_{k_1}={\mu }_{k_1}{N}_1$ , and we get,

 

$f_{k_1}={\mu }_{k_1}{m}_{1}g\cos \theta$ 

 

Rewrite the equation  $m_{1}a=m_{1}g\sin \theta -T-f_{k_1}$ as follows, and we get,

 

 $$\begin{gathered} m_{1}a=m_{1}g\sin \theta -T-{\mu }_{k_1}{m}_{1}g\cos \theta \\ T=m_{1}g\left(\sin \theta -{\mu }_{k_1}\cos \theta \right)-m_{1}a \end{gathered}$$

 

The forces acting on mass  $m_2$ are shown in the following figure.

Similarly, from the above figure, the relation of forces on the horizontal direction is given by,

 

 $T=m_{2}a-m_{2}g\left(\sin \theta -{\mu }_{k_2}\cos \theta \right)$ 

 

Here $m_2$  is the mass of the second block and $f_{k_2}$  is the kinetic friction due to the second block.

 

The tension force acts on the mass   and   are equal.

 $$\begin{gathered} m_{1}g\left(\sin \theta -{\mu }_k\cos \theta \right)-m_{1}a=m_{2}a-m_{2}g\left(\sin \theta -{\mu }_{k_2}\cos \theta \right) \\ \left(m_1+m_2\right)a=m_{1}g\left(\sin \theta -{\mu }_{k_1}\cos \theta \right)+m_{2}g\left(\sin \theta -{\mu }_{k_2}\cos \theta \right) \\ a=\frac{m_{1}g\left(\sin \theta -{\mu }_{k_1}\cos \theta \right)+m_{2}g\left(\sin \theta -{\mu }_{k_2}\cos \theta \right)}{\left(m_1+m_2\right)} ..........\left(1\right) \end{gathered}$$

Substitute  $9.8m/s^2$ for g ,  4 kg for $m_1$ ,  8 kg for $m_2$ ,  $30°$ for $\theta$ ,  0.25 for ${\mu }_k$ , and  0.35 for  ${\mu }_{k_2}$ in equation (1).

$$\begin{gathered} a=\frac{\left(9.8m/s^2\times 4kg\right)\left(\sin 30°-0.25\times \cos 30°\right)+\left(9.8 m/s^2\times 8 kg\right)\left(\sin 30°-0.35\times \cos 30°\right)}{4kg+8kg} \\ =\frac{11.1129+15.4363}{12}m/s^2 \\ =2.212 m/s^2 \end{gathered}$$  

 

Therefore, the acceleration of each block is $2.212m/s^2$ .

Step 4: Find the tension in the string (b)

Substitute $9.8m/s^2$  for g , 4kg  for $m_1$ ,  $30°$ for $\theta$ ,  0.25 for ${\mu }_{k_1}$ ,  0.35 for ${\mu }_{k_2}$ , and $2.212m/s^2$  for a  in the equation $T=m_{1}g\left(\sin \theta -{\mu }_{k_1}\cos \theta \right)-m_{1}a$ , and we get,

 $$\begin{gathered} T=\left(9.8m/s^2\times 4kg\right)\left(\sin 30°-0.25\times \cos 30°\right)-4 kg\times 2.212m/s^2 \\ =2.265 N \end{gathered}$$ 

 

Therefore, the tension in the string is 2.265 N .

Step 5: Find the change in condition when the positions of the blocks are reversed (c)

The force on   is given by,

 $F_1=m_{1}g\left(\sin \theta -{\mu }_k \cos \theta \right) .......\left(2\right)$

 

Substitute  $9.8 m/s^2$ for g ,  4 kg for $m_1$ ,  $30°$ for $\theta$ ,  0.25 for ${\mu }_{k_1}$ , and $2.212m/s^2$  for a  in the equation (2).

$$\begin{gathered} F_1=\left(4 kg\right)\left(9.8m/s^2\right)\left[\sin 30°-0.25\cos 30°\right] \\ =11.11N \end{gathered}$$  

 

The force on  $m_2$ is given by,

$F_2=m_{2}g\left(\sin \theta -{\mu }_{k_2}\cos \theta \right) .......\left(3\right)$ 

 

Substitute $9.8m/s^2$  for g ,  8 kg for $m_2$ , $30°$  for $\theta$ , 0.35  for ${\mu }_{k_1}$ , and  2.212$m/s^2$ for a  in the equation (2), and we get,

 

$$\begin{gathered} F_2=\left(8kg\right)\left(9.8m/s^2\right)\left[\sin 30°-0.35\cos 30°\right] \\ =15.44N \end{gathered}$$ 

 

If the position of the block is revered, the acceleration can be calculated as follows.

 

Calculate the acceleration of mass ${\mathrm{m}}_1$  as follows.

$$\begin{gathered} a_1=\frac{F_1}{m_1} \\ =\frac{11.11N}{4Kg} \\ =2.78m/s^2 \end{gathered}$$ 

 

Calculate the acceleration of mass $m_2$  as follows.

$$\begin{gathered} a_2=\frac{F_2}{m_2} \\ =\frac{15.44N}{8 kg} \\ =1.93 m/s^2 \end{gathered}$$ 

 

Therefore, if the positions of the block are reversed, then the acceleration of the blocks is $2.78m/s^2$ , and $1.93m/s^2$ .