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# In the arrangement shown in the figure, the mass of the body A is n=4 times that of body B. The height h=20cm. At a certain instant, the body B is released and the system is set in motion. What is the maximum height, the body B will go up? Assume enough space above B and A sticks to the ground.(A and B are of small size)(g=10 m/sec^2)Can anyone please help me in this question?

Shivam Chopra
45 Points
6 years ago
Here, for A, FBD =
( T= Tension force by the rope, 4g = weight of block)
As, it will go down, so, equation becomes
4g – 2T = 4a, where, 4a = net force on block A, a = acceleration of block
For B, FBD =

g = weight of bock.
As, it will go upwards, so, equation becomes
T – g = a
Solving both equations, we get a = g/3 = 10/3 ms-2
As, the acceleration is same, it applies for same time & initial velocity of blocks is zero;
So, the distance covered by both blocks is same i.e. 20 cm
Riya
8 Points
6 years ago
I cant view the image
Shivam Chopra
45 Points
6 years ago
Sorry about that. I am resending the picture
Sandeep
20 Points
3 years ago
Above given seems to be incomplete, when block A reaches the ground block B will have acquired some velocity and it cover some more distance
Harsh
15 Points
one year ago
After solving u can get that block B will be displaced by 40 cm.
But during this interval it will gain some velocity and after 40 cm the string will slack and Tension will be zero.
Then apply law of motions and get the answer.