IIT JEE , AIEEE Forum - Physics , Mechanics

shashankparewar

in the figure the spring is compressed by a mechanical means .the compression is just sufficient that when the spring is at its maximum elongation then 2kg block just leaves surface.find the
normal force applied by the ground on the 2kg block as the function of time.

feynmann

Is it 3.5 ( 1 + cos wt ) ?

shashankparewar

exactly i dont remember the soln but give me the method.

gaurav3sharma

SHOULDN'T IT BE MAXIMUM COMPRESSION( INSTEAD OF ELONGATION)???? BECAUSE YOU SAY THAT THE SPRING IS BEING COMPRESSED.........PLEASE CLEAR THIS..

tanveergill

Consider the time when the spring is in max elongation to be t=0.

Assume max elongation as "L"

2 Kg block can never leave contact with ground as maximum force acting on it just tends to lift it.

Therefore equation of shm for 3 Kg block: Lsin(wt+

/2)

Spring force = K* Lsin(wt+

/2)

On the 2 Kg block the following forces act:

1. Spring Force T in upward direction

2. Normal reaction N in upward direction

3. mg in downward direction

The 2 Kg block is in equilibrium hence:

T + N = mg

K* Lsin(wt+

/2) + N = 2g -1

Now K=mw²=3w²

K*L = 2g because at L elongation, 2 kg block tends to leave ground

3w²* L = 2g

L = 2g/3w²

Put value of L and K in 1

2gsin(wt+

/2) + N = 2g

N = 2g(1 - sin(wt+

/2))

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