# How to solve a spring question?

What is the difference between the distance a spring is stretched using $F=kx$ formula vs the distance using the work formula? They seem to give different answers.

Doesn't distance stretched using the $F=kx$ formula indicate maximum distance stretched by applying a certain force? This confusion is indicated in this spring question in my textbook:

A spring with k = 53N/m hangs vertically next to a ruler. The end of the spring is next to the 19-cm mark on the ruler. If a 2.0-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?

I used hookes law, $F=kx$, and found x=0.37m. Adding in the initial mark of 19cm I found 56cm. Unfortunately it says this is wrong answer. Answer is found by using work potential energy formula!

• Could you add some more details about the whole setup? What is pulling in the spring to make it elongate? Where do you have the force $F$, which turns out to give you the certain elongation $x$ og 37 cm? Nov 30, 2016 at 17:34
• Given the way this is written, I can't tell what the book is even asking (looks like it is stating a couple facts to consider). Could you elaborate more on what the book is asking? Nov 30, 2016 at 17:37
• Sorry. Yep I was missing part of question. Its edited now.
– TLo
Nov 30, 2016 at 18:01

You seem to have a confusion over zero force and zero velocity. When you extend a spring by a distance x, the restoring (pulling) FORCE exerted by it on you is kx. Now at an elongation of 0.37m, as you calculated, the NET FORCE acting on the ball is zero, since the upward pull of the spring will be exactly equal and opposite to the downward force of gravity.

Situation 1: Instead of releasing the ball when it is next to the 19cm mark as mentioned in the question, hold the ball in your hand and slowly pull the ball down by 0.37m. Now release the ball. You will notice that the ball will stay where it is. This is because you released the ball with no initial speed and the net force on the ball is zero. Nothing happens.

Situation 2: The question mentions that you release the ball at the 19 cm mark. Here the net force on the ball is NOT zero. It will be in the downward direction,causing it to accelerate downwards. Now as the ball falls, it will start to gain speed, since before reaching an extension of 0.37m, the net force on the ball is downwards (gravity exceeds the spring force). By the time the ball has fallen by 0.37m, it would have gained a finite velocity. Although the net force on the ball is zero, its velocity is surely not. So it will continue to move downwards, past the equilibrium mark, since nothing can stop its motion abruptly. What happens after it crosses the extension of 0.37m? Now the net force on the ball is UPWARDS (spring force exceeds gravity). So the ball will start losing speed and will eventually come to a stop. This is when the extension in the spring will be maximum.

Bit the story is not over yet. The spring which has extended by quite a bit will pull the ball upwards again, with an venture greater force. Overall, the ball will undergo simple oscillations about the equilibrium position.

Finally, notice that the total energy if the spring, ball and earth system will be conserved. The spring will have maximum potential energy when the kinetic energy if the ball is zero(zero speed).

To summarize, a zero force implies zero acceleration and not zero velocity.

What you have calculated is the equilibrium position $P$ of the mass, which it reaches when it eventually comes to rest. However, before that happens the mass oscillates around $P$. If there is no energy loss it will continue oscillating forever.

It is not necessary to use the conservation of energy principle. (I think you mean this rather than the work-energy theorem.) The oscillations are symmetric about $P$. The point of release is at maximum displacement above $P$. The point of maximum displacement below $P$ is the same distance below it. So the lowest point will be $2 \times 0.37m$ below the point of release.

Using conservation of energy, the elastic energy stored in the spring at the lowest point equals the gravitational PE lost by the mass :
$\frac12kx^2=mgx$
$x=2(mg/k)$
which gives the same answer. If you apply both methods correctly you will get the same answer.