# Compression of spring when an object of given mass is placed on it

A question in my textbook says - A block of mass $m$ is released from rest onto an ideal non deformed spring from a negligible height. Neglecting air resistance, find compression $x$ of the spring. I tried using two approaches for this Q but I can't figure out which one if right or if both are wrong.

1st -

Using Hooke's Law. $$mg -kx = 0$$ $$x = (mg) / k$$

2nd - Using Work-Energy theorem. Considering the Gravitational potential energy at the lowest point of the compression to be 0 and since change in KE is 0.

$$mgx -(kx^2)/2 = 0$$ $$mgx = (kx^2)/2$$ $$x = (2mg)/k$$

Is any of my approach correct? If yes then why is the other one wrong? Or are both completely wrong?

• Well this isn't really a 'check my work' problem according to me. I only showed my work so as to show that I was not simply looking for someone give me the answer and had actually given a lot of thought to the question. – MayankJain Sep 30 '15 at 5:56
• Oh. Didn't even think of that . Thanks a ton! @Timaeus, so is there any way to find the maximum compression at using the force approach? – MayankJain Sep 30 '15 at 6:09

Both are right. The first approach gives the compression where the net force on the object is zero. The second approach gives the compression when the velocity of the object is zero. When the block falls on the spring, it oscillates between $x=\frac{2mg}{k}$ and $x = 0$. Since the spring is ideal and the air resistance is negligible, this oscillation does not die down and so the question is wrong. But if the oscillation dies down eventually, the spring comes to rest at $x = \frac{mg}{k}$.

• Oh. Okay. Got it. So is there any way to find the maximum compression using the force approach? – MayankJain Sep 30 '15 at 6:11
• Since it's an SHM, the magnitude of force at one extremum must be the same at the other extremum. Since you already know one extremum and the net force acting on it there, you can find the other extremum. – sarat.kant Sep 30 '15 at 6:16
• Hmm.. I'm not really familiar with SHM at the moment. So in that case, the work approach is the way to go, yeah? – MayankJain Sep 30 '15 at 6:19
• Yes it is. Just ask your teacher about the correctness of this question. – sarat.kant Sep 30 '15 at 6:20

The work-energy theorem is certainly the easiest way to do the problem, but you can also solve it by calculating the force.

In any situation where you need to calculate the response of an object to a force you use Newton's second law. This tells us (after a minor rearrangement):

$$\frac{d^2x}{dt^2} = \frac{F}{m} \tag{1}$$

In this case the force on the mass has two parts. There is a downwards force of $F_{gravity} = mg$ due to gravity, and an upwards force due to the spring. Assuming the spring obeys Hooke's law then the force when the spring is compressed a distance $x$ is given by:

$$F_{spring} = -kx$$

The net force on the mass if $F_{gravity} + F_{spring}$, and putting this into equation (1) we get:

$$\frac{d^2x}{dt^2} = \frac{mg - kx}{m}$$

If you solve this differential equation you get an equation for $x$ as a function of time. It's then straightforward to find the maximum value of $x$.

The equation is a lot easier to solve than it looks, because it's the equation for a simple harmonic oscillator in disguise. However if you haven't done simple harmonic motion yet this is probably not a route you want to pursue.

• I think you meant Fgravity + Fspring? (Sorry for the lack of MathJax, it doesn't seem to work on my mobile device) – MayankJain Sep 30 '15 at 7:21
• @MayankJain: oops yes. – John Rennie Sep 30 '15 at 7:44

according to me you have erred in assuming that change in KE is 0 as the block will have some velocity when it is compressed.since total mechanical energy remains constant loss in P.E will result in gain of K.E