Question

001. Two blocks A-20 kg, 8-15 kg, are hanging on either side of a pulley of mass 1 kg. Calculate the linear acceleration of the blocks.
Opst AO100/53 m/s?
B. O 100/71 m/s​

Answer 1

Answer:

I don't know the answer thanks for asking

Answer 2

Answer: Acceleration of the blocks is $\bold{\frac{100}{71}\ m/s^2}$.

Given: Two blocks A-20 kg, B-15 kg and pulley of mass 1 kg.

To Find: Linear acceleration of the blocks.

Step-by-step explanation:

Step 1: Let R be the radius of the pulley and $T_1$ and $T_2$ be the tensions in the left and right portions of the string. Contrary to the rules of motion, the tension of the string on either side of the pulley is the same. In this instance, the pulley turns and the string does not slide. There will be a difference in the tension on the pulley's two sides. The pulley will experience an angular acceleration due to the various stresses.

Step 2: Let, $m_1 =20kg$

$m_2 =15kg$

$M = 1kg$

Let a be the acceleration of blocks.

For the blocks (linear motion)

$T_1 -m_1 g=m_1 a ...(i)\\\\m_2g-T_2 =m_2 a ...(ii)$

For the pulley (rotation)

Net torque $\tau_{net }=I\alpha$

$T_2R-T_1R=\frac{1}{2} MR^2\alpha$

Step 3: The linear acceleration of the blocks is the same as the tangential acceleration of any point on the circumference of the pulley which is Rα

Dividing Eq.(iii) by R and adding to Eqs. (i) and (ii), we get

$m_2g-m_1g=m_2a+m_1a+\frac{M}{2} R\alpha \\\\m_2g-m_1g=(m_2+m_1+\frac{M}{2}) a\\\\a=\frac{m_2-m_1}{m_2+m_1+\frac{M}{2} } \ g\\\\a=\frac{20-15}{20+15+\frac{1}{2} } \times 10\\\\a=\frac{100}{71}\ m/s^2$

Hence, acceleration of the blocks is $\frac{100}{71}\ m/s^2$.

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