Question

Q. Two blocks of masses 4 kg and 10 kg are connected with a massless string and a massless spring in between them. String goes over a massless and frictionless pulley as shown in figure. Initially the spring is unstretched. When 4 kg block is released from rest, minimum normal contact force between 10 kg block and the surface in contact with it will be (Take g = 10m/s?)

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Answer 1

Concept:-The normal force will be minimum, only when,the spring force or expansion in the string will be maximum...

Let the expansion in the string be x,due to the tension created by blocks.

As there is no external forces,and we know that the sum of internal forces always eqauls to zero,,so work done by these forces will be zero..

So,

work done in expanding the string- change in potential energy of 5kg block=0

work done in expanding the string=stored potential energy in string$=\frac{1}{2}k {x}^{2}$

Change in potential energy of block of 4kg$= mgx$

$\frac{1}{2}k {x}^{2} - mgx = 0 \\ \frac{1}{2}k {x}^{2} = mxg \\ kx = 2mg$

we know that the normal force on any block kept at earth surface is eqaul to it's weight ,,

But in this case normal force is decreased due to spring force in upward direction,so,

$normal + kx = mg \\ nomal \: force = 100 -80 = 20$

This will be the minimum normal force between the block and the surface in contact......

Answer 2

꧁ Correct answer is 60 N ꧂