Question

29. Figure shows a block of mass 2m sliding on a block of mass m. Find the acceleration of each block. (All
surface are smooth)​

Answer 1

Answer:

$a = \frac{g}{6}=\frac{10}{6} =\frac{5}{3} m/s^2$

Explanation:

1) We have,

Two blocks of mass 'm' and '2 m' are released .

Let the acceleration of mass '2m' be 'a' downward.  

Writing Force equation on mass 'm'.

$2mgsin(\theta)-T=2ma$

and

Same on mass 'm'.

$T-mgsin(\theta)=ma$

2) Now,

Adding above two equations,

$mgsin(\theta)=3ma\\ \\=>a=\frac{g}{3}sin(\theta)$

Since,$\theta = 30^{\circ}$

Therefore,$a=\frac{g}{3}sin(30^{\circ})=\frac{g}{3}*\frac{1}{2}=\frac{g}{6}$

Hence, acceleration of mass '2m' is g/6 units in downward along the incline.

Acceleration of mass 'm' is g/6 up the incline.

Answer 2

Answer:

Explanation:

acceleration = mgsin30 / 3m

acceleration = g/6