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

To solve this problem, we'll determine the time taken for the box with a mass of 8 kg to strike the ground when starting from rest. We need to consider the system of two blocks connected by a string over a pulley.

Initially, let's denote:

• Mass of block A (m₁) = 2 kg
• Mass of block B (m₂) = 8 kg
• Distance to fall = 20 cm = 0.2 m
• Acceleration due to gravity (g) = 10 m/s²

The blocks will accelerate under the effect of gravity due to the difference in mass.

Step 1: Calculate net acceleration of the system

The net force (F_net) can be calculated as:

F_{\text{net}} = m_2 \cdot g - m_1 \cdot g

In terms of acceleration (a):

(m_1 + m_2)\cdot a = (m_2 - m_1) \cdot g

Solve for acceleration (a):

a = \frac{(m_2 - m_1) \cdot g}{m_1 + m_2}

Substitute the given values:

a = \frac{(8 - 2) \cdot 10}{2 + 8} = \frac{60}{10} = 6\, \text{m/s}^2

Step 2: Determine the time taken to fall 20 cm

Use the equation of motion:

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

Since the system starts from rest, u = 0.

Substitute values:

0.2 = 0 + \frac{1}{2} \cdot 6 \cdot t^2

0.2 = 3 \cdot t^2

t^2 = \frac{0.2}{3} = \frac{1}{15}

t = \sqrt{\frac{1}{15}} \approx 0.2582

Since the total distance is double due to the pulley, calculate for 40 cm:

t_{\text{total}}^2 = \frac{2 \times 0.2}{6} = \frac{0.4}{6} = \frac{1}{15} \times 2 \approx 0.4

Conclusion

Thus, the time taken for the 8 kg block to strike the ground is approximately 0.4 seconds.