Blocks of mass m and 3m are attached to the two ends of a light inextensible string and passed over a light pulley. Find the acceleration of the lighter block(take g=10m/s^2)?

Question

Answer

Problem Statement: Two blocks of masses m and 3m are attached to the ends of a light inextensible string and passed over a light pulley. Find the acceleration of the lighter block (take g=10m/s^2).

Solution:

Step 1: Free Body Diagrams of the Blocks
The free body diagrams of the two blocks are shown below:

Block of mass m: Tension T acts upward, Weight W = mg acts downward

Block of mass 3m: Tension 3T acts upward, Weight W' = 3mg acts downward

![image.png](attachment:image.png)

Step 2: Equations of Motion
Since the two blocks are connected by an inextensible string, the acceleration of both blocks is the same and can be represented by a.

For the block of mass m: T - mg = ma

For the block of mass 3m: 3T - 3mg = 3ma

Step 3: Elimination of Tension T
We can eliminate tension T by solving the two equations simultaneously.

From the first equation, T = ma + mg

Substituting T in the second equation, we get: 3(ma + mg) - 3mg = 3ma

Simplifying, we get: 2ma = mg

Therefore, a = g/2

Step 4: Final Answer
The acceleration of the lighter block is g/2 or 5 m/s^2.

Answer

5 m/s2

Answer

5m/s^2 for both the blocks

Answer

Total force / total mass of the system..

Answer

5

Answer

Suppose the mass 3m is on the right side of pulley and mass m is on the left side. Total force acting = 30-10N =20NThis force pulls the heavier mass 3m downwards and pulls the lighter mass m upwards. Since string and pulley is massless both will have same acceleration which is 20/(1+3) =5m/s^2

Answer

10m/s^2... Because the pulley is at rest

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