Question

A pop-up toy consists of a mass m stuck to the top of a light spring of natural length `l_0`and spring constant k. The spring is compressed to length `l_1` when the pop-up is stuck to the ground. To what height above the ground does it reach?

A pop-up toy consists of a mass m stuck to the top of a light spring of natural length `l_0` and spring constant k. The spring is compressed to length `l_1` when the pop-up is stuck to the ground. To what height above the ground does the bottom of the unstretched spring jump to when it is smoothly released?

Answer 1

The height is `h=k/(2mg)(l_0-l_1)(l_0-l_1-1)`

Explanation:

The mass of the toy is `=mkg`

The spring constant is `=kNm^-1`

The compression of the spring is

`Deltax=(l_0-l_1)m`

The energy stored in the spring is

`E=1/2kDeltax^2=1/2k(l_0-l_1)^2`

This energy will be transformed into potential energy

`PE=mgh`

The height is `=hm`

The acceleration due to gravity is `=gms^-2`

Therefore,

`PE=E`

`mg(l_0+h-l_1)=1/2k(l_0-l_1)^2`

`mgh+mg(l_0-l_1)=1/2k(l_0-l_1)^2`

`h=k/(2mg)(l_0-l_1)^2-(l_0-l_1)=k/(2mg)(l_0-l_1)(l_0-l_1-1)`

The height is `=k/(2mg)(l_0-l_1)(l_0-l_1-1)`

Answer 2

`{(l0-l1)(kl0-kl1-2mg)}/(mg)`

Explanation:

Let's assume that the point where the spring is attached,the potential energy is zero.

So,the potential energy of the mass is `mgl0`

When the spring got compressed to a length of `l1`,it caused in two things,

One is decrease in the potential energy of the mass i.e new potential energy of the mass is `mgl1` and the spring due to compression achieved an amount of elastic potential energy i.e `(1/2)k((l0)-(l1))^2`

So we can tell this potential energy which got stored in the spring was equals to that amount of decrease in potential energy of mass m , which occurred as it came to `l1` from `l0`

So,we can write,
`mg((l0)-(l1))= (1/2)k((l0)-(l1))^2`
Or, `l0 - l1 = (2mg/k)`

Now,when the spring will be smoothly released, if the bottom point of spring goes up to height `h` above the ground,total potential energy of mass m becomes `mg((l0)+h)`

So,from conservation of energy we can write,
Total energy of the system due to compression of the spring = total energy of the system after reaching the maximum height
[ Note one thing we consider here,that is the fully stretched spring (that is at its original length) has no energy]

Therefore,
`mg(l0+h)` = `(1/2)k((l0)-(l1))^2 + mg l1`

So we get, `h = {(l0-l1)(kl0-kl1-2mg)}/(mg)`