1.

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.

Answer»


Solution :(a) At POINT B,
`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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