Question

If m1 is moving with acceleration 'a' in the direction shown in the figure. Then the acceleration of M is

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Answer 1

The correct answer will be ∴ a2=F−M1​a1/M2.

Explanation:

• Let consider x be the flexibility in the spring.  

• The equation of motion of the blocks are:

• F−kx=M1a1.........................(1)

• kx=M2a2.........................................(2)

• Adding these equations in the formula we will get F=M1a1+M2a2

• By further solving it

• ∴ a2=F−M1​a1/M2

• hence, option D is correct.

Answer 2

Given:

"M1" is moving with "a" (acceleration).

To Find:

Acceleration of M = ?

Solution:

Let the spring's flexibility be "x".

Now,

The block's equation of motion will be:

⇒  $F-kx=M1\times a1$ ...(equation 1)

and,

⇒  $kx=M2\times a2$ ...(equation 2)

On adding the above "equation 1" and "equation 2", we get

⇒  $F-kx+kx=M1a1+M2a2$

⇒  $F=M1a1+M2a2$

Therefore,

⇒  $a2=\frac{F-M1a1}{M2}$

So that the acceleration of M will be "$a2=\frac{F-M1a1}{M2}$".