Respuesta :
Answer:
[tex]0.1\; {\rm J}[/tex], assuming that the wire acts like an ideal spring.
Explanation:
Let [tex]k[/tex] denote the spring constant of this wire under the assumptions. Hooke's Law relates the restoring force [tex]F[/tex] from the spring to the displacement [tex]x[/tex] of the spring:
[tex]F = -k\, x[/tex].
In this question, it is given that [tex]F = (-200)\; {\rm N}[/tex] when [tex]x = 1\; {\rm mm} = 10^{-3}\; {\rm m}[/tex]. Note that the restoring force is equal in magnitude to the external force, but negative since the two forces are opposite in direction.
Rearrange this equation to find the spring constant [tex]k[/tex]:
[tex]\begin{aligned}k &= -\frac{F}{x}\end{aligned}[/tex].
When the displacement of spring of spring constant [tex]k[/tex] is [tex]x[/tex], the elastic potential energy stored in the spring would be:
[tex]\begin{aligned}(\text{EPE}) &= \frac{1}{2}\, k\, x^{2}\end{aligned}[/tex].
Substitute in [tex]k = (-F/x)[/tex] to find [tex](\text{EPE})[/tex] in terms of [tex]F[/tex] and [tex]x[/tex]:
[tex]\begin{aligned}(\text{EPE}) &= \frac{1}{2}\, k\, x^{2} \\ &= \frac{1}{2}\, \left(- \frac{F}{x}\right)\, x^{2} \\ &= -\frac{F\, x}{2} \\ &= -\frac{(200\; {\rm N})\, (10^{-3}\; {\rm m})}{2} \\ &= 0.1\; {\rm J}\end{aligned}[/tex].
The elastic energy stored in a wire when it is stretched by a weight of 200 N to extend by 1 mm is 0.1 J. The energy is calculated using the formula E = 1/2 * F * x. the correct answer is option C. 0.1 J.
The student's question asks about the elastic energy stored in a wire when it is stretched by a weight. To find the elastic energy stored in a wire due to stretching, you can use the formula for the potential energy stored in a spring (or an elastic material), which is E = 1/2 * k * x2, where truek is the spring constant and x is the extension of the spring.
However, the spring constant is not given here, but we know the force applied (200 N) and the extension caused by the force (1 mm or 0.001 m). Since we're dealing with a linear elastic material, and assuming Hooke's Law applies, we know that the force is proportional to the extension, which allows us to rewrite the elastic potential energy formula as E = 1/2 * F * x, where F is the force and x is the extension.Therefore, we can calculate the elastic energy as E = 1/2 * 200 N * 0.001 m, which gives us 0.1 J. So, the correct answer is C. 0.1 J.
