Question

Two blocks A and B of mass $10kg$ and $40kg$ are connected by an ideal string a shown in the figure. Neglect the masses of the pulleys and effect of friction in the pulley and between the blocks and the inclines (wedges is fixed)

• A. The acceleration of block A and B is $13.2m{s}^2,6.6m{s}^2$
• B. The acceleration of block A and B is $12.2m{s}^2,5.6m{s}^2$
• C. The acceleration of block A and B is $11m{s}^2,3m{s}^2$
• D. The acceleration of block A and B is $23m{s}^2,43m{s}^2$

Answer 1

The correct option is A The acceleration of block A and B is $13.2m{s}^2,6.6m{s}^2$

As per constant relation if acceleration of block $B$ is ${}^{\prime \prime }{a}^{\prime \prime }$ then

acceleration of block $A$ is $"2a"$

now as per force equation of $B$

$\begin{array}{c}{m}_{b}g\sin 45-2T=m_{b}a\\ T-m_{a}g\sin 45=m_a\times 2a\end{array}$

now solving above two equations for acceleration of the blocks

$\begin{array}{c}{m}_{b}g\sin 45-2m_{a}g\sin 45=\left(m_b+4m_a\right)\times a\\ a=\frac{\left(m_b\sin 45-2m_a\sin 45\right)g}{m_b+4m_a}\end{array}$

now plug in all values

$\begin{array}{c}a=\frac{\left(20\sqrt{2}-10\sqrt{2}\right)9.8}{40+40}\\ a=6.6m/s^2\end{array}$

So, acceleration of block $B$ is $6.6m/s$ and acceleration of block $A$ is $13.2m/s^2$