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A 500 g block rests on a frictionless horizontal table at a distance of 400 nm from a fixed pin O. Then block is attached to pin O by an elastic cord of constant k = 100 N//m & of underformed length 900 mm. If the block is set in motion perpendiculariy as shown |
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Answer» the speed V to be set initially for which the distance from O to the block P will REACH the maximum value of 1.2 m is 4.5 m/s `v = 3u` also by energy conservation  `(1)/(2) xx 0.5 xx v^(2) = (1)/(2) xx 100 xx (0.3)^(2) + (1)/(2) xx 0.5 xx u^(2)` `rArr (v^(2))/(4) = (9)/(2) + (u^(2))/(4)` `rArr (9u^(2))/(4) = (9)/(2) + (u^(2))/(4) rArr (8u^(2))/(4) = (9)/(2)` `rArr u = sqrt(9)/(4) = (3)/(2)` So `v = 3u = 3 xx 1.5 = 45` (C ) `a_(n) = (u^(2))/(r) "" rArr r = (u^(2))/(a_(n))` `(a_(n) = (K_(h))/(m) = (100 xx 0.3)/(0.5) = 60m//s^(2))` `rArr r = ((1.5)^(2))/((60)) = (2.25)/(60) = 0.0375m` `= 3.75cm` |
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