Q93P - University Physics with Modern Physics

Question

Two objects, with masses  5.00 kg and 2.00 kg , hang 0.600 m  above the floor from the ends of a cord that is  6.00 m long and passes over a frictionless pulley. Both objects start from rest. Find the maximum height reached by the  2.00 kg object.

Step-by-Step Solution

Answer

The object can rise to a maximum height of 1.46 m .

Step 1: Identification of the given data

The given data can be listed below as:

• The mass of the first object is $m_1=5.00 kg$ .
• The mass of the second object is $m_2=2.00 kg$ .
• The hanging distance of the objects from the floor is $s=0.600 m$ .
• The length of the cord is $l=6.00 m$ .

Step 2: Significance of free-body diagrams

The diagram of an isolated part of a system, in which all the force acting on that part are depicted, is known as a free body diagram. They are very helpful in calculating the value of forces that are unknown.

Step 3: Determination of the maximum height reached by the first object

The equation of the acceleration of the second object is expressed as:

 

  $a=\frac{\left(m_1-m_2\right)g}{\left(m_2+m_1\right)}$

 

Here,  is the acceleration of that object, $m_2$ is the mass of the second and $m_1$ is the mass of the first object. g is the acceleration due to gravity.

 

For the given values, above equation becomes-

 $$\begin{gathered} a=\frac{\left(5.00 kg-2.00 kg\right)\left(9.8 m/s^2\right)}{\left(5.00 kg+2.00 kg\right)} \\ =\frac{\left(3.00 kg\right)\left(9.8 m/s^2\right)}{\left(7.00 kg\right)} \\ =\frac{\left(29.4 kg.m/s^2\right)}{\left(7.00 kg\right)} \\ =4.2 m/s^2 \end{gathered}$$

The equation of the time taken by the second object to move a distance :

 $s=ut+\frac{1}{2}a{t}^2$ 

 

Here,  s is the hanging distance of the objects from the floor, u  is the initial speed of that object and  is the time taken by that object to reach to the distance.

 

As the object was at rest initially, the initial speed of that object is zero.

 

For the given values, above equation becomes-

 

 $$\begin{gathered} 0.6 m=\left(0\right)t+\frac{1}{2}\left(4.2m/s^2\right){\left(t\right)}^2 \\ 0.6 m=\left(2.1 m/s^2\right){\left(t\right)}^2 \\ t^2=0.28 s^2 \\ t=0.53 s \end{gathered}$$

 

The equation of the final velocity of the second object is expressed as:

$u^2=v^2-2g{s}_1$ 

  

 Here, $s_1$ is the final velocity of the second object.

 

For the given values, above equation becomes-

 

$$\begin{gathered} v=0+\left(4.2m/s^2\right)\left(0.53s\right) \\ =\left(4.2m/s^2\right)\left(0.53s\right) \\ =2.25 m/s \end{gathered}$$ 

 

The equation of the initial height reached by the second object is expressed as:

 $u^2=v^2-2g{s}_1$

  

 Here, $s_1$ is the initial height reached by the second object. 

 

For the given values, above equation becomes-

 

 $$\begin{gathered} 0^2={\left(2.25m/s\right)}^2-2\left(9.8m/s^2\right)s_1 \\ 5.0625m^2/s^2=\left(19.6m/s^2\right)s_1 \\ {s}_1=0.25 m \end{gathered}$$

 

The equation of the maximum height reached by the second object is expressed as:

 

 $h_{max}=2s+s_1$ 

 

Here, $h_{max}$ is the height reached by the second object.

 

For the given values, above equation becomes-

 

 $$\begin{gathered} h_{max}=2\left(0.6 m\right)+0.25 m \\ =1.2 m+0.25 m \\ =1.46 m \end{gathered}$$

 

Thus, the highest,  2.00 kg object can go, is 1.46 m .