Answer:
Work done is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Explanation:
The work done by the spring is the same as the potential energy stored in the spring.
So that,
work done = potential energy = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
where k is the spring constant of the material of the spring, and x is the compression.
When the spring is compressed by x;
work done = [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
When the spring is compressed by 2x;
work done = [tex]\frac{1}{2}[/tex] k[tex](2x)^{2}[/tex]
= [tex]\frac{1}{2}[/tex] k(4[tex]x^{2}[/tex])
= 2k[tex]x^{2}[/tex]
Therefore,
The work done the second time more than the first = 2k[tex]x^{2}[/tex] - [tex]\frac{1}{2}[/tex] k[tex]x^{2}[/tex]
= [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex]
The work done the second time more than the first is [tex]\frac{3}{2}[/tex]k[tex]x^{2}[/tex].
Spring work is equivalent to the effort done to extend the spring, Work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
Spring work is equivalent to the effort done to extend the spring, which is dependent on both the spring constant k and the distance stretched.
The potential energy stored as a result of the deformation of an elastic item, such as spring stretching, is referred to as elastic potential energy.
Work done by spring = potential energy
[tex](PE)_{spring }= \frac{1}{2} Kx^2[/tex]
Case 1
spring is compressed by x
[tex](PE)_1{spring }= \frac{1}{2} Kx^2[/tex]
Case 2
spring is compressed by 2x
[tex]\rm (PE)_2{spring }= \frac{1}{2} K(2x)^2\\\\\rm (PE)_2{spring }= \frac{1}{2}\times 4K(x)^2\\\\\rm (PE)_2{spring }= 2K(x)^2[/tex]
The difference in the potential energy is found by;
[tex](PE)_2-(PE)_1=2Kx^2-\frac{1}{2} Kx^2=\frac{3}{2} Kx^2[/tex]
Hence spring work did you do the second time than the first time will be [tex]\rm \frac{3}{2} Kx^2[/tex].
To learn more about the spring work refer to the link;
https://brainly.com/question/3317535
