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 ?

Options

A. particle should be dropped from a height of 3.96m to cause a compression of 0.04 m.
B. Conservation of energy will be followed by the particle.
C. particle should be dropped from a height of 5.42m to cause a compression of 0.04 m.
D. None of these.

Answer

A, B

Explanation

When a particle is dropped from a height h and the spring is compressed by x, then loss in potential energy of the particle `= mg (h + x)`.
The gain in elastic potential energy of the spring `=(1)/(2) kx^(2)` where k is the spring constant.
From the principle of conservation of energy, we have `mg(h+x)= (1)/(2) kx^(2)`.
In the first case, `m=0.1 kg, h=0.24 m, x=0.01 m`.
`therefore 0.1 g (0.24 +0.01)= (1)/(2) k (0.01)^(2) " "...(i)`
In the second case, `m=0.1 kg, h=h_(2) m (say) , x=0.04 m`
`therefore 0.1 g (h_(2) +0.04) = (1)/(2) k (0.04)^(2) " "...(ii)`
Dividing (ii) by (i) we get, `(h_(2) + 0.04)/(0.24 + 0.01) = ((0.04)/(0.01))^(2)`
or, `h_(2) + 0.04 =16 xx 0.25`
or, `h_(2) =3.96` m
So, the particle has to be dropped from a height of 3.96 m.