Question

A $40\;cm$ long spring becomes $47\;cm$ when a load of $10\;N$ hangs from its free end. Determine the force constant of the spring.

Answer 1

Hint:When a force is applied by the weight of the load which is attached to the spring there is an extension in its length. This extra length of the spring once the load is hung on it needs to be calculated first. The formula for the force due to a load on a spring given by Hooke’s law is applied and the value of the force constant is found out.

Formula used:
$F = - k \times x$
where, $F$ is the restoring force, $k$ is called the force constant and $x$ is the displacement or extension of the spring.

Complete step by step answer:
Let us consider a spring of negligible mass to which a load is applied which is of weight $10\;N$ while the length of the string is $40\;cm$ as per the data obtained from the above question. The length of the spring when no load is applied is $40\;cm$ and the length after the load is changed is $47\;cm$.

Since there is an extension that is produced, we calculate this extra length of the extended spring. This is obtained by subtracting the length of the spring when there was no load attached to it from the length of the extended spring.
$ \Rightarrow 47 - 40 = 7\;cm$
This is now converted into meters.
We know that,
$1\;m = 100\;cm$
By applying unitary method we get,
$ \Rightarrow 7\;cm = 0.07\;m$

Now, the following formula for finding the force constant is applied:
The formula used is in accordance to Hooke’s law (by only taking its magnitude and neglecting the negative symbol) is:
$F = k \times x$
Here, the force of the spring is called the restoring force because objects that have a property of oscillation or motion try to go back to their original position to attain stability.
Since we need to find the force constant given the force of the load applied, F and the extension x and by rearranging the terms of the above formula we get:
$k = F/x$
The values which are given are substituted in the above formula.
$ \Rightarrow k = 10 \div 0.07$
$ \Rightarrow k =142.85\;N/m$
The answer can be written in terms of $N/cm$ as well,
$ \Rightarrow k =1.4285\;N/cm$
This can be rounded off to 3 significant figures
$\therefore k = 1.43\;N/cm$

Hence the force constant obtained is $1.43\;N/cm$.

Additional information: The force of the spring which has been extended is said to be negative as this signifies that the force is in the opposite direction to the displacement in the string after a load is applied to it from its unstretched position. The force of the spring, called the restoring force, is also known as tension in a string. A string of negligible mass is said to have a constant tension, say, $T$ throughout the string.

The spring constant is a measure of the stiffness of the spring which means that how much it gets tends to restore to its original position. The stiffness is the value of spring constant $k$.When the spring is pulled downwards by the load, the spring retaliates by exerting an upward force to balance out the force that is applied on it in accordance with Newton's law.

Note:There is a possibility to take the negative sign of the formula for Hooke’s law into consideration which is wrong since the sign only denotes the direction of pull of the string which is downwards. The magnitude should not be negative.