for the arrangemnet shown in thefigure,
the spring is initially compressed by 3cm.
when the spirng is released the block collides with the wall and rebounds to compress the spring again.
q>> if the time starts at the instant when spring is eleased, find the minimum time afetr which the block becomes stationary..
[plz answer lucidly... coz this is something i dunno ... ]
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3 Answers
106initially the spring is compressed by 3cm.. so initial PE = 104*(0.03)2/2 = 4.5 J
KE of the system after the block leaves the spring = 1.v2/2
Now, from conservation of energy .. KE gained = PE lost
==> v2 = 9
==> v = 3 m/s
Now .. time taken to reach the initial position where it left the spring = 2L/v = 2*0.04/3 = 0.0267 s
now.. for time taken for the block .. to stop. .. i.e. it will undergo SHM with amplitude 3cm.. at time period.. 2Ï€√1/104 s = Ï€/50 s
Now.... the time taken by block .. to leave the spring initially is equal to 1/4th of an oscillation and similarly... to reach the stopping point again is 1/4th of the oscillation ...
So. total time in oscillating = T/2 = π/100 s = 0.0314s
So total time taken = 0.0581 s
3theres no non conservative force in system so it only comes momentarily to rest [1][]1
21aSIS : "Now .. time taken to reach the initial position where it left the spring = 2L/v = 2*0.04/3 = 0.0267 s"
bUT if the picture shown is in free state then t = D/v wich u hav used will bcum (2*0.04 + 0.03)
As the s[ring is compressd by 3 cm
AND CAN U PL. TELL me in ur workin u took T/2 can u trace that path!!!
