Question

Two blocks are kept on a smooth fixed slopes as shown. 2 kg block goes down with an acceleration of 2 $m/s^2$. The value of m (in kg) is :-

• A. 0.4
• B. 0.6
• C. 0.8
• D. 1

Answer 1

Answer: (C)

0.8

Answer 2

Here's the breakdown of the solution:

1. Forces on the 2 kg block:

   • Component of gravity down the slope: $m_1 g \sin(\theta_1) = 2 \cdot 10 \cdot \sin(37^\circ) \approx 2 \cdot 10 \cdot 0.6 = 12 \, \text{N}$
   • Tension (T) acting upwards

   Equation of motion: $12 - T = m_1 a = 2 \cdot 2 = 4 \, \text{N}$ Therefore, $T = 12 - 4 = 8 \, \text{N}$

2. Forces on the mass m block:

   • Tension (T) acting upwards
   • Component of gravity down the slope: $m g \sin(\theta_2)$

   Assuming $\theta_2 = 53^\circ$ (since the slopes are at $90^\circ$ and one is at $37^\circ$), then $\sin(53^\circ) \approx 0.8$.

   Equation of motion: $T - m g \sin(\theta_2) = m a$ $8 - m \cdot 10 \cdot 0.8 = 2m$ $8 - 8m = 2m$ $10m = 8$ $m = 0.8 \, \text{kg}$

   Therefore, the value of m is 0.8 kg.