Question

Both the blocks shown in figure have same mass 'm'. All the pulleys and strings are massless.

• A. Acceleration of block A is $\frac{2g}{5}$.
• B. Acceleration of block A is $\frac{g}{5}$.
• C. Acceleration of block B is $\frac{g}{5}$.
• D. Tension in the string attached with block A is $\frac{3mg}{5}$.

Answer 1

Answer: (A), (C), (D)

Acceleration of block A is $\frac{2g}{5}$, Acceleration of block B is $\frac{g}{5}$, Tension in the string attached with block A is $\frac{3mg}{5}$.

Answer 2

Let $a_A$ be the downward acceleration of block A and $a_B$ be the downward acceleration of block B. Let $T$ be the tension in the string.

Equation of motion for block A: $mg - T = ma_A$.

Equation of motion for block B: $mg - 2T = ma_B$.

Kinematic constraint: $a_A + 2a_B = 0$, so $a_A = -2a_B$.

Let $a_B' = -a_B$ be the upward acceleration of block B. Then $a_A = 2a_B'$.

Substitute $a_B = -a_B'$ and $a_A = 2a_B'$ into the equations of motion:

$mg - T = m(2a_B')$

$mg - 2T = m(-a_B')$

From the first equation: $T = mg - 2ma_B'$.

Substitute into the second equation: $mg - 2(mg - 2ma_B') = -ma_B'$

$mg - 2mg + 4ma_B' = -ma_B'$

$-mg = -5ma_B'$

$a_B' = \frac{g}{5}$.

So, upward acceleration of block B is $\frac{g}{5}$.

Downward acceleration of block A is $a_A = 2a_B' = 2\left(\frac{g}{5}\right) = \frac{2g}{5}$.

Tension $T = mg - 2ma_B' = mg - 2m\left(\frac{g}{5}\right) = mg - \frac{2mg}{5} = \frac{3mg}{5}$.

Checking the options:

(A) Acceleration of block A is $\frac{2g}{5}$. This is the magnitude of acceleration of A, and also the downward acceleration. Correct.

(C) Acceleration of block B is $\frac{g}{5}$. This is the magnitude of acceleration of B, and also the upward acceleration. Correct.

(D) Tension in the string attached with block A is $\frac{3mg}{5}$. This is correct.