Question

In the arrangement shown in the Fig., the block of
mass m = 2 kg lies on the wedge of mass M = 8 kg.
Find the initial acceleration of the wedge if the surfaces
are smooth.
M M
60°​

Answer 1

Hence the value of acceleration is A= 3√3 g / 23 ms−1

Explanation:

Acceleration with the incline:

a = A ( 1 +  cos θ ) 3 A / 2 (Since θ = 60∘)  

F.B.D of M

Equation of motion:

For mass "m" : mg √ 3 / 2 + m A × 1 / 2 − T = m 3 / 2 A  ------(1)

N + m A √3 / 2 =  mg  1/2  -----(2)

For M: T + N √3 / 2 = M A  ----(3)

From equations. (1), (2) and (3) we can see that

A= 3√3 g / 23 ms−1

Hence the value of acceleration is A= 3√3 g / 23 ms−1

Answer 2

Answer:

Explanation:

Hence the value of acceleration is A= 3√3 g / 23 ms−1

Explanation:

Acceleration with the incline:

a = A ( 1 +  cos θ ) 3 A / 2 (Since θ = 60∘)  

F.B.D of M

Equation of motion:

For mass "m" : mg √ 3 / 2 + m A × 1 / 2 − T = m 3 / 2 A  ------(1)

N + m A √3 / 2 =  mg  1/2  -----(2)

For M: T + N √3 / 2 = M A  ----(3)

From equations. (1), (2) and (3) we can see that

A= 3√3 g / 23 ms−1

Hence the value of acceleration is A= 3√3 g / 23 ms−1