Question

A system consists of two identical masses A and B of mass m each connected to ends of a massless spring of force constant k as shown in figure
The minimum force F applied vertically downward on A, such that on its release, B will leave the
floor.
(1) mg
(2) 2mg
(3) 3mg
(4) 4mg​

Answer 1

Answer:

OK ans is exclaimed

Explanation:

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Answer 2

$In\ equilibrium, F +mg =k\Delta x \ (1)$for block B to be lifted up.

$kx\geq mg \rightarrow x_{min}=\frac{mg}{k}$  (2)

From the conservation of energy,

$[mg(x+\Delta x)-0]+[\frac{1kx^{2} }{2} -\frac{1k(\Delta x^{2}) }{2} ]=0$

$mg(x+\Delta x) +\frac{1k}{2} (x+\Delta x)(x-\Delta x)=0$

$(x+ \Delta x)[mg+\frac{(x-\Delta x)k}{2} ]=0$

$\rightarrow\frac{(\Delta x-x)k}{2} =mg$

$\rightarrow \Delta x=\frac{2mg}{k} +\frac{mg}{k} =\frac{3mg}{k}\\\\\frac{F+mg}{k} =\frac{3mg}{k} (from\ 1)\\F=2mg$

Hence, the minimum force required for B to leave the floor is 2mg.

The correct option is (2) 2mg.