Problem 50

From an orkut discussion

Post 1

-------

}

}====>spring

}A

!'''''''!-------------------()-====>pulley

---horizontal surface] !

--------------------------] !

--------------------------] !

--------------------------] !''''''!====>block B

fig. shows 2 blocks A and Beach having a mass of 320g connected by a light string passing over a smooth light pulley.the horizontal surface on which the block A can slide is is smooth.the block A is attached to a springof spring constant 40N/m whose other end is fixed to a support 40cm above the horizontal surface.initially the spring is vertical and unstretched when the system is rreleased to move.find the velocity of the block A at the instant it breaks off the surface below it.g=10m/s^2

Post 2

hey the answer is 1.5m/s^2

i started off the problem by assuming that there will be no extension in the spring until the block A travels a distance L0sinteeta[L=the length of the spring, teeta the angle the spring makes with the vertical]

then i applied conservation of energy[before and after extension]

then i got struck!!!!!!!!!

is this correct what i did?

if not please let me know how to start of with this problem...

Post 3

the exact ans is 1.54m/s..........

when d block has moved a dist x then the vertical component of force of spring balance mg i.e. 3.2N and we calculate extention in spring from there ...... extention comes out to be 0.1m ,,....then apply cons of energy between initial and final pos........

E1= -mgh

E2= 1/2mv^2 -mg(h+x)

E1=E2

h is the length of hanging block which will cancel out x being = Ltan (theeta)

and theeta is calculated when we know extention of spring

........so v=1.54m/s

## 1 comment:

how does it becomes h+k but not

h2+k2

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