Question

The two blocks A and B of equal mass are initially in contact when released from the rest on the inclined plane . O be the angle of inclined plane . The coefficient of friction b/w the inclined plane and A and B are u1 and u2 respectively (u1 is greater than u2 ) . The block B is placed besidethe Block A.Then what is the common acceleration of two blocks .

Answer 1

Here is your answer:

Explanation:

Force due to lower block in the downward direction

mg ( Sin θ - µ1 Cos θ)

Force due to upper block in the downward direction

mg ( Sin θ - µ2 Cos θ)

Total force in the downward direction

(m + m) a = mg ( Sin θ - µ1 Cos θ) +  mg ( Sin θ - µ2 Cos θ)

  2ma     = 2mg Sinθ - mg ( µ1 + µ2 ) Cosθ

Common acceleration of the blocks = g sinθ - g/2 ( µ1 + µ2 ) cosθ

Answer 2

The common acceleration of block A and block B is g (sin θ - μ₁ cos θ)

Explanation:

The force acting on block A on the downward direction is:

ma = mg sin θ - μ₁ mg cos θ

⇒ ma = mg (sin θ - μ₁ cos θ)

The force acting on block B on the downward direction is:

ma = mg sin θ - μ₂ mg cos θ

⇒ ma = mg (sin θ - μ₂ cos θ)

Since, we need to find the common acceleration,

ma + ma = (mg (sin θ - μ₁ cos θ)) + (mg (sin θ - μ₂ cos θ))

2 ma = 2 mg sin θ - 2 mg cos θ (μ₂ - μ₁)

a = g sin θ - g cos θ (μ₂ - μ₁)

∴ a = g (sin θ - (μ₂ - μ₁) cos θ)