IIT JEE , AIEEE Forum - Physics , Mechanics

perin

HCV part 1 page 135 Q.50
figure shows two blocks A and B each having a mass 320 g connected by a light string passing over a smooth light pulley. the horizontal surface on which block A can slide is smooth. the block A is attached to a spring with K = 40 N/m whose other end is fixed to a support 40 cm above the horizontal surface. find the velocity of the block A at the instant it breaks off the surface below it. take g = 10 m/s ²

varun.tinkle

50)
the natural length is 0.4 and let the xtended length be 0.4+x
mg=kxcos@
cos@=mg/kx
which is also equa to 0.4/0.4+x
solving which we get x as 0.1
mgh=1/2mv^2+1/2mv^2+kx^2
since velocity of both the blocks will be the same
solving it we wll get v as 1.5

ramyani

Waterdemon solved it.

The Energy is stored in the following three forms:

(a)Elastic Potential Energy of the Spring:

(Stretched by a length of "dL")

= (1/2)*K*(dL)²

(b)K.E of Block A = (1/2)mv²

(c)K.E of Block B = (1/2)mv²

Now Applying the law of conservation of energy,

mgx = (1/2)k(dL)² + (1/2)mv²+ (1/2)mv²

v² = gx - (k/2m)(dL)² ..........(1)

Considering the Equilibrium, of Block A at position A :

R + FCos@ = mg ...........(2)

For a Spring,

F = K(dL) and for Breaking off R = 0.

Substituting the value of R and F in Eq.(2)

K(dL)Cos@ = mg ..........(3)

From the Geometry of the problem ,

dL = (L/Cos@) - L = L{ (1/Cos@) - 1 }

Substituting in Eq.(3)

K{ L [ (1/Cos@) - 1 ]Cos@ = mg

On solving we get,

1 - Cos@ = mg/KL

Cos@ = [1-(mg/KL)]

Cos@ = 1 - (0.32*10/40*0.40) = 4/5

and dL = 0.4(5/4-1)

dL = 0.1 m from above.

Now x=L(tan@) = 0.4*(0.3/0.4) = 0.3m.

Subsitituting the values in eq.(1)

v² = 10*0.3 (40*(0.1)²/2*0.32)

v² = 3 - 0.64

v² = 2.36

v = 1.5 m/s.

Hope you find it useful.

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Cheers!!!!!!!!!!!!!!

hsb

x= dist moved by a bfore losing contact

decrease in pe of system= mgx

loss in pe =gain in epe = ke of 2 blocks

mgx = 1/2 ka^2 + mv^2

calc v^2 = gx - k/2m*a^2

when A loses contact vert eqm fcos

= mg

f = k a

solving a cos theta =.08

L/a+L = COS THETA cos theta = .1

calculate v

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