Question

Two blocks $A$ and $B$ of mass $10kg$ and $40kg$ are connected by an ideal string as shown in the figure. Neglect the masses of the pulleys and effect of friction in the pulleys and between the blocks and the inclines. Then

• A. The acceleration of block $A$ is $\frac{19}{\sqrt{2}}m{s}^{-2}$
• B. The acceleration of block $B$ is $\frac{50}{2\sqrt{2}}m{s}^{-2}$
• C. The tension in the string is $\frac{125}{\sqrt{2}}N$.
• D. The tension in the string is $\frac{1500}{\sqrt{2}}N$.

Answer 1

The correct option is A The acceleration of block $A$ is $\frac{19}{\sqrt{2}}m{s}^{-2}$

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

acceleration of block $A$ is ${}^{\prime }2a^{\prime }$

Now as per force equation of $B$

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

now solving above two equations for acceleration of the blocks

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

now plug in all values

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

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