A pop-up toy consists of a mass m stuck to the top of a light spring of natural length l_0and 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?
from Isaac physicsfrom Isaac physics

2 Answers
Jan 14, 2018

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)

Jan 14, 2018

{(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)