Answer:
2[tex]x_{1}[/tex], The spring would compress twice the length of [tex]x_{1}[/tex].
Explanation:
The Spring force is an application of Hooke's law, and using the law of conservation of energy in a spring system, we can find the relationship between the compression length and velocity.
For a spring system, the conservation of energy can be applied using the formula below;
[tex]\frac{1}{2} kx_{1} ^{2} = \frac{1}{2} mv^{2}[/tex] ........................................1
Where k is the spring constant
[tex]x_{1}[/tex] is the distance
m is the mass and
v is the velocity
We have to make the distance the subject formula to know how it affects the velocity;
[tex]x_{1} ^{2} = (\frac{m}{k} )v^{2}[/tex]
[tex]x_{1} = \sqrt{(\frac{m}{k} )}v[/tex] .....................................2
from the equation, it shows that the distance ([tex]x_{1}[/tex]) is directly proportional to the velocity (v). This implies that when the distance increases the velocity also increases.
x ∝ v
Therefore when the initial speed is doubled to 2 v it will amount a double increase in the distance which is 2[tex]x_{1}[/tex]
If the initial speed of the box were doubled, the spring would get compressed a distance [tex]x_2 = 2\,x_1[/tex].
Here, while the box slides on the surface it possesses kinetic energy.
This kinetic energy is converted into elastic potential energy and is stored in the spring.
So, according to the law of conservation of energy, we can write;
[tex]\frac{1}{2}mv^2= \frac{1}{2}(kx_1)^2[/tex]
On rearranging the above equation. we get;
[tex]v^2 =\frac{k}{m} (x_1)^2 \\\\\implies v=\sqrt{\frac{k}{m}}\,\,x_1 \\\\\implies v \propto x_1[/tex]
So, we get that the compression is directly proportional to the initial speed.
If the initial speed is doubled, the compression will also be doubled
i.e.; [tex]x_2 = 2\,x_1[/tex]
Learn more about the law of conservation of energy here:
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