Question

a pulley is assumed massless and frictionless and rotates freely about its axle. the blocks have masses m1=40g and a block m2= 20g, and block m1 is pulled to the right by a horizontal force magnitude f=0.03n. find the magnitude of acceleration of a block m2 and the tension in the cord.....?

Answer 1

The magnitude of acceleration of a block m₂ is 0.05 m/s² and the tension in the cord is 0.01 N.

Given:

mass of block 1, m₁ = 40 gm = 40×10⁻³ kg

mass of block 2, m₂ = 20 gm = 20×10⁻³ kg

Applied force, F = 0.03 N

Calculation:

Consider the free-body diagram of the system as shown below. Using Newton's second law of motion we get:

F = ma

where F is the applied force

            m is the total mass of the system

            a is the acceleration of block 2 (as it is pulled by horizontal force)

From the above equation we get:

0.03 N = (m₁+m₂) a

⇒ a = (0.03 N) / (m₁+m₂)

⇒ a = (0.03 N) / (40×10⁻³ kg + 20×10⁻³ kg)

⇒ a = 0.5 m/s²

Now, from the free-body diagram of block 2 as shown in figure 3, we get:

Balancing the forces along the horizontal:

∑Fₓ = 0

∴ T = m₂ a

where T is tension in the string

           a is the acceleration of block 2

Applying values in the above equation we get:

T = (20×10⁻³ kg) × (0.5 m/s²)

  = 0.01 N

Therefore, the acceleration of block 2 due to the applied horizontal force is 0.5 m/s² and the tension in the cord is 0.01 N.

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