Question

The spring in the figure stretches from 10 cm to 22 cm when a force of 4N is applied. Calculate:
• its force constant k
• its total length when a force of 6 N is applied.

Answer 1

Answer:

the formula f=kx

Explanation:

so f=6

Delta x=12cm

f/x=k

k=4/12

k=1/3

Answer 2

Formula to be used:-

• Force exerted by a spring with a spring constant k when it is compressed or extended by a distance $\sf \Delta x$ is given by $\sf F_{(spring)} = -k\Delta x$. This is also known as Hooke's law. Here, $\sf \Delta x = (x_f - x_i),$ where $\sf x_{i}$ is the initial compression/elongation of the spring, and $\sf x_{f}$ is the final compression/elongation of the spring.

Answer:-

The force to be applied to pull the spring = - Force exerted by spring

$\sf F = - F_{(spring)}$

$\longrightarrow \sf F =-( -k\Delta x)$

$\sf \longrightarrow F = k\Delta x$

$\sf \longrightarrow F = k(x_f - x_i) \quad \quad \dots (1)$

FInding spring constant 'k' :-

Considering that the spring is pulled from its natural position, which means $\sf 10\;cm = 0.1\;m$ is its natural length, and after being pulled $\sf 22\;cm = 0.22\;m$, the elongation will be $\sf (0.22-0.10)\;m = 0.12\:m$.

So,

In the first case,

• Here, $\sf x_i = 0\;m$ as we have considered that we are pulling from natural length. And so, the initial compression/elongation will be 0
• And, $\sf x_f = 0.12\;m$, as we found out the elongation to be $\sf 0.12\;m$.

It is given that the force required to be applied is $\sf F = 4\;N$.

Putting this in (1),

$\sf  F = k(x_f - x_i)$

$\longrightarrow \sf 4 = k(0.12 - 0)$

$\longrightarrow \sf 4 = k\times 0.12$

$\longrightarrow \sf k = \dfrac{4}{0.12}$

$\longrightarrow \underline{\underline{\sf{\green{ k = 33.33\;N/m} }}}$

Finding total length when a force of 6 N is applied:-

Let the total length be $\sf l\; metres$.

We are considering that we are pulling this spring from its natural length, that is $\sf 0.1\;m$, and so the final elongation will be $\sf l - 0.1\;m$

So,

• Here, $\sf x_i = 0\;m$ as we have considered that we are pulling from natural length. And so, the initial compression/elongation will be 0
• And $\sf x_f = l - 0.1\;m$, as we found out the final elongation to be $\sf l - 0.1\;m$.

Putting this in (1),

$\sf \longrightarrow F = k(x_f - x_i)$

$\longrightarrow \sf 6 =33.33\big\{(l - 0.1)-0\big\}$

$\longrightarrow \sf \dfrac{6}{33.33} = l - 0.1$

$\longrightarrow \sf 0.18= l -0.1$

$\longrightarrow \sf l = 0.28\;m$

$\longrightarrow \underline{\underline{\sf{ \green{l = 28\;cm }}}}$