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In the figure sown in the concept, find (a) Equilbrium extension x_(@) (=AB) (b) Maximum extension x_(m) (=AC) (c)Maximum speed at point B. |
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Answer» `F_("NET") =0` `rArr kx_(0) =mg` `rArr x_(0) mg/K` .From A to C `(v_(A)=v_(C) =0)` Decreasing in gravitational potential energy = incrasing in spring energy. `:. mgx_(m)=-1/2Kx_(m)^(2) , AC=x_(m)` `rArr x_(m)=2mg/K` (c) FORM A to B Dereasing ingravitational potential energy = increasing in (spring potential energy + kinetic energy) `rArr mg x_(0) =1/2 Kx_(0)^(2) +1/2mv_("max")^(2)` Substituting the value of `x_(0) =mg/K` in the above eqation, we get `v_("max")=(SQRT(m/K))g`. |
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