1.

A block of mass 'm' initially at rest is dropped from a height 'h' into a spring whose constant is k. The maximum distance through which the spring is compressed is :

Answer»

`(2mgh)/(k)`
`(2mg(h+k))/(3)`
`(mg-sqrt(m^(2)G^(2)+2mghk))/(2k)`
`(mg+sqrt(m^(2)g^(2)+2mghk))/(k)`

Solution :Here P.E. of fall`=mg (h+x)=1/2kx^2`
`2mgh+2 mgx= kx^2`
or`kx^2-2 mgx-2 mgh=0`
`x=(2mg PM sqrt(4m^2g^2+4kxx2 mgh))/(2k)`
=`(mg pm sqrt(m^2g^2+2mghk))/(k)`
(Take +ve SIGN)


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