Question

The boxes of masses 2 kg and 8 kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8 kg to strike the ground starting from rest. (use g = 10 m/s\(^2\)): 

 

• A. 0.2 s
• B. 0.34 s
• C. 0.25 s
• D. 0.4 s

Hint:

In pulley problems, carefully establishing the constraint relation between accelerations is the most critical first step. A common mistake is to assume accelerations are equal. For a movable pulley like this, remember that for every 'x' distance it moves, the free end of the string moves '2x'.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We have a pulley system with two masses, 8 kg and 2 kg. The 8 kg mass is attached to a movable pulley. We need to find the time it takes for the 8 kg mass to fall a distance of 20 cm, starting from rest.
Step 2: Key Formula or Approach:
1. Constraint Relation: Determine the relationship between the accelerations of the two masses based on the string length.
2. Newton's Second Law: Apply \( F_{\text{net}} = ma \) to each mass to set up equations of motion.
3. Kinematics: Use the equation \( s = ut + \frac{1}{2}at^2 \) to find the time.
Step 3: Detailed Explanation:
Constraint Relation:
Let the downward acceleration of the 8 kg mass (\(m_1\)) be \(a_1\). Let the upward acceleration of the 2 kg mass (\(m_2\)) be \(a_2\).
If the 8 kg mass moves down by a distance \(x\), the length of the string on both sides of its pulley is released. This total length \(2x\) is pulled by the 2 kg mass. So, the 2 kg mass moves up by a distance \(2x\).
Differentiating twice with respect to time, we get the relation between accelerations: \( a_2 = 2a_1 \).
Let \( a_1 = a \), then \( a_2 = 2a \).
Equations of Motion:
Let \(T\) be the tension in the string attached to the 2 kg mass. The movable pulley supporting the 8 kg mass is pulled up by two segments of this string, so the total upward force on it is \(2T\).
For the 8 kg mass (\(m_1\)):
The net downward force is \( m_1g - 2T \).
\( m_1g - 2T = m_1a_1 \)
\[ 8(10) - 2T = 8a \] \[ 80 - 2T = 8a \quad \text{(Equation 1)} \] For the 2 kg mass (\(m_2\)):
The net upward force is \( T - m_2g \).
\( T - m_2g = m_2a_2 \)
\[ T - 2(10) = 2(2a) \] \[ T - 20 = 4a \quad \text{(Equation 2)} \] Solving for Acceleration 'a':
From Equation 2, we get \( T = 4a + 20 \).
Substitute this expression for \(T\) into Equation 1:
\( 80 - 2(4a + 20) = 8a \)
\( 80 - 8a - 40 = 8a \)
\( 40 = 16a \)
\[ a = \frac{40}{16} = \frac{5}{2} = 2.5 \, \text{m/s}^2 \] This is the downward acceleration of the 8 kg block.
Calculating the Time:
The 8 kg block needs to fall a distance \( s = 20 \, \text{cm} = 0.2 \, \text{m} \).
It starts from rest, so initial velocity \( u = 0 \).
Using the kinematic equation: \( s = ut + \frac{1}{2}at^2 \)
\( 0.2 = (0)t + \frac{1}{2}(2.5)t^2 \)
\( 0.2 = 1.25 t^2 \)
\( t^2 = \frac{0.2}{1.25} = \frac{20}{125} = \frac{4}{25} \)
\[ t = \sqrt{\frac{4}{25}} = \frac{2}{5} = 0.4 \, \text{s} \] Step 4: Final Answer:
The time taken for the 8 kg box to strike the ground is 0.4 s.