Question

Two blocks A and B of masses $m_1$ = 2kg and $m_2$ = 8kg respectively arc allowed to move without friction. Block A is on block B and block B slides on smooth fixed inclined plane as shown in the figure. The angle of inclination of inclined plane is 45°. (Take g = 10 m/s²)

The acceleration of the block B with respect to the ground is

• A. $\frac{5\sqrt{3}}{7} m/s^2$
• B. $\frac{50\sqrt{2}}{9} m/s^2$
• C. $10\sqrt{2} m/s^2$
• D. $16\sqrt{5}m/s^2$

Answer 1

Answer: (B)

$\frac{50\sqrt{2}}{9} m/s^2$

Answer 2

The acceleration of block B with respect to the ground is given by the formula

$a_B = \frac{(m_1+m_2)g \sin\theta}{m_2 + m_1 \sin^2\theta}$

Substituting the given values, $m_1 = 2$ kg, $m_2 = 8$ kg, $\theta = 45^\circ$, $g = 10 m/s^2$.

$a_B = \frac{(2+8) \times 10 \times \sin 45^\circ}{8 + 2 \times \sin^2 45^\circ} = \frac{10 \times 10 \times \frac{1}{\sqrt{2}}}{8 + 2 \times (\frac{1}{\sqrt{2}})^2} = \frac{\frac{100}{\sqrt{2}}}{8 + 2 \times \frac{1}{2}} = \frac{\frac{100}{\sqrt{2}}}{8+1} = \frac{100}{9\sqrt{2}} = \frac{100\sqrt{2}}{18} = \frac{50\sqrt{2}}{9} m/s^2$.