Question

# A block m2 is loaded onto another block m1 placed on a spring of stiffness k as shown in the figure. If block m2 is suddenly removed, then magnitude of acceleration of block m1 just after the removal of m2 will be:

A
(m2m1g)
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B
m1m2m1+m2g
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C
m1+m2m1m2
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D
(m1m2g)
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Solution

## The correct option is A (m2m1g)Considering the equilibrium of system, just before removal of m2. FBD before removal of m2: Here, x= compression in spring kx=(m1+m2)g…(i) ∴ x=(m1+m2)gk Spring force will remain as it is after removal of m2, because spring will not change its length instantaneously. FBD after removal of m2: Applying Newton's 2nd law on block m1 in direction of acceleration: kx−m1g=m1a ⇒(m1+m2)g−m1g=m1a (from (i)) ∴a=(m2m1g) Hence, block m1 will move with acceleration (m2m1g) upwards.

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