263(1): H=L+(mg + √m2g2+2mgLk)/k
(2): V=√2gL+mg2/k
Swarna Kamal Dhyawala how plz explain??
Upvote·0· Reply ·2014-07-01 10:58:32
A block of mass m is attached to one end of a long elastic rope. The other end of the elastic rope is fixed to the roof of a building. Initially the block is in contact with the roof at the point where the rope is fixed and is allowed to fall freely from rest. The unstretched length of the rope is L and has a force constant k.
(i) If the block just reaches the floor find the height of the room.
(ii) what is the maximum speed of the block during this drop.
(iii) The time taken during the drop before coming to rest for first time.
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2 Answers
263a=(mg-kx)/m............(1)[where x is the extension of the string after reaching it's natural length]
Now at the point where the string regains it's natural length velocity is √2gl.See from the equation (1) that the acceleration is positive if x<=mg/k.
now in equation (1) writing a as v(dv/dx),we get
v(dv)=(g-kx/m)(dx).................(2)
Integrating this equation with the boundary condition on v from V to √2gl and on x from mg/k to 0 we get V2=2GL + mg2/k.This is the maximum velocity of the block because till x=mg/k acceleration is positive.
Now after x=mg/k acceleration is negative.The block's velocity reduces from V to 0.Now it is already given that the block touches the ground when it's velocity is 0.So integrate again equation (2) with the conditions that :
Lower limit of v=V
Upper limit of v=0
Lower limit of x=mg/k
Upper limit of x=h,where h is measured from the point where the string attains it's natural length.
A quadratic equation comes in term of h.Solve it to reach the answer.
Also plz tell the answer to question (3).
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