A massless platform is kept on a light elastic spring as shown in figure. When a small stone of mass 0.1 kg is dropped on the pan from a height of 0.24 m, the spring compresses by 0.01m. From what height should the stone be droppped to cause a compression of 0.04m in the spring ?

Explanation

Here, `m=0 kg, h=0.24m, x=0.01m. `
Clearly, loss in P.E. of stone `=mg(h+x), ` and gain in elstic P.E. of spring `=(1)/(2)kx^(2)`.
`:. Mg(h+x)=(1)/(2)kx^(2)` …(i)
Let `h^(')` be the height through which stone should be dropped to cause a compression of `x^(')=0.04m`.
`:. mg(h^(')+x^('))=(1)/(2)kx^(').^(2)` ...(ii)
Dividing (i) by (ii), we get
`(h+x)/(h^(')+x^('))=((x)/(x^(')))^(2)=((0.01)/(0.04))^(2)=(1)/(16)`
`h^(')x^(')=16h+16x`
`h^(')+0.04=16xx0.24+16xx0.01`
`=3.84+0.16=4.00`
`h^(')=4.00-0.04=3.96m`