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Figure shows a block A on a smooth surface attached with a spring of force constant k to the ceiling. In this state spring is in its natural length l . The block A is connected with a massless and frictionless string to another identical mass B hanging over a light and smooth pulley. Find the distance moved by A before if leaves contact with the ground. |
Answer» Solution : Due to weight of BLOCK B, it moves down and PULLS the block A now as block A and B move, spring gets stretched and becomes inclined as its lower end is attached to the block A. It will break off from the GROUND below it when the VERTICAL component of the spring force on block A will balance its weight mg Let it happens when A moves by a distances as shown in figure. At this instant let the spring be inclined at an angle with the vertical If the stretch in the spring at this instant is x, then it is given as `x = l sec theta - l ` or ` x = l ( sec theta - 1)` ........... (1) If mass A breaks off from ground below it , we have ` kx cos theta = mg ` From equation (1) and (2) substituting the value ofx, we get `kl( sec theta - 1) cos theta = mg or kl(1- cos theta ) = mg or cos theta =1 -(mg)/(kl) or tan theta =([k^(2)l^(2) (kl - mg)^(2) ]^(1//2))/(kl-mg)` At this instant the distance travelled by mass A and B is given by ` s=l tan theta (or) s=l([k^(2)l^(2)(kl-mg)^(2)]^(1//2))/(kl-mg)` |
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