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 :

Options

A. `(2mgh)/(k)`
B. `(2mg(h+k))/(3)`
C. `(mg-sqrt(m^(2)g^(2)+2mghk))/(2k)`
D. `(mg+sqrt(m^(2)g^(2)+2mghk))/(k)`

Answer

d

Explanation

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)