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Figure shows tow blocks A and B, each having a mass of 320 g connected by a light string passing over a smooth light puley. The horizontal surface on which the block A can slide is smooth. The blcok A is attached to a spring of lconstant 40 N/m whose other end is fixed to a support 40 cm the horizontal surface. Initially, the sprng is vertical and unstretched when the system is released to move. Find the velocity of theh block A at teh instnt it breaks of the surface below it. Take g=10 m/s^2. |
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Answer» `h=40cm=0.4m,g=10m/sec^2` From the free body diagram, `kxcostheta=mg` (when the BLOCK breaks of R=0)` `RARR costheta=(mg)/(kx)` `rarr 0.4/(0.4+x=3.2/(40x)` `rarr 16x=3.2x+1.28` `rarr x=0.1m` So, `s=AB=SQRT((h+x^2)-h^2)` `=sqrt((0.5)^2-(0.4)^2)`=0.3m` Let the velocity of tebody at B be v change in K.E. work done (for the system) `(1/2mu^2+1/2mv^2)=-1/2kx^2+mgs` `rarr (0.32)xxv^2` `=-(1/2)xx40xx(1.0)^2+(0.2)xx10xx(0.3)` `rarr v=1.5m/sec` `From the ure `/_\l=h(sectheta-1)`........i From the energy conservation `mgs=2[1/2mv^2]+1/2K/_\l^2` ` mgh than theta =mv^2+1/2kh^2(sectheta-1)^2`.....ii From the law of motion at break of `mgkh(sectheta-1)costheta` `rarr 1-costheta=mg/KH` `rarr costheta=1-mg/kh` `or costheta=(kh-mg)/kh` or `costheta=(kh-mg)/kh` `=(40xx0.4-0.32xx10)/(40xx0.4)` `=0.8` PUTTING the value of `theta` in equation i, `0.32xx10xx0.4xx0.75` `=0.32v^2+1/240xx(0.4)^2(1.25-1)^2` `rarr 0.96=0.32v^2+0.2` `rarr `0.32v^2=0.72` `rarr v=1.5m/sec |
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