Question

Two masses 2 kg and 4 kg are connected at the two ends of light inextensible string passing over a frictionless pulley. If the masses are released, then what is the acceleration of block 2 kg? [Take g = 10 m/s2 ]



Choose one:

3.00 m/s2

6 m/s2

3.33 m/s2

6.33 m/s2

Answer 1

Given,

Two masses 2 kg and 4 kg are connected at the two ends of light inextensible string passing over a frictionless pulley as shown in picture.

Now,

let the common acceleration of both objects be a and tension of string be T

For object of mass of 4 kg

4g>T

Therefore, resultant force will try move mass of 4kg in downward direction,

So,

4g - T = 4a .................................(i)

For object of mass of 2kg,

T>2g

Therefore, resultant force will try move mass of 2kg in upward direction,

T - 2g = 2a .................................(II)

Adding Equation (i) and (ii),

We get,

4g - 2g = 6a

2g = 6a

2*10 = 6a

6a = 20

a = 10/3

a = 3.33 m/sec²

Answer 2

» smaller mass = m1 = 2kg

» larger mass = m2 = 12 kg

» Tension = T

______________

» For mass m1

=> T - m1g = ma ____eq(1)

» For mass m2

=> m2g - T = m2a. ____eq(2)

_____________

» By eq(1) + eq(2)

=> (m2 - m1)g = (m1 + m2)a

=> a = (m2 - m1)g/(m1 + m2)

=> a = (4-2)×10/(4+2)

=> a = 2×10/6 = 20/6

=> a = 3.33 ms^-2

_______________[ANSWER]

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