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Answer :
(a)The spring constant will be 1,329 N/m.
(b)The length of the spring will be 44.92 cm.
What is the spring constant?
Spring constant is defined as the ratio of force per unit displaced length. The spring force is balanced by the weight;
The given data in the problem is;
L₁ is the long spring is hung vertically from a ceiling= 31.0 cm
L₂ = 36.9 cm
m is the mass= 8.00 kg
The net elongation of the spring is
x= 36.9-31 =5.9 cm = 0.059 m
The force acted on the spring due to which the elongation is done;
F=mg
F= 8 × 9.81
F=78.4 n
From the Hooke's law the spring constant is found as;
[tex]\rm K = \frac{F}{x} \\\\ \rm K = \frac{78.4}{0.059} \\\\ \rm K = 1,329 \ N/m[/tex]
Hence the spring constant will be 1,329 (in N/m).
(b) The length of the spring will be 44.92 cm.
For the weight of F=185 N is hung from the spring, the elongation is:
[tex]\rm x= \frac{185}{1,329} \\\\ \rm x=0.10392 m = 13.92 \ cm[/tex]
The length of spring is;
[tex]\rm x_1 = x_0+x \\\\ \rm x_1 = 31+13.92 \\\\ \rm x_1=44.92\ cm[/tex]
Hence the length of the spring will be 44.92 cm.
To learn more about the spring constant reference to the link;
https://brainly.com/question/4291098
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Answer:
The length of the spring is 44.92 cm
Explanation:
Hooke's Law
Suppose a spring of constant k and natural length x0. If a force F is applied to the spring and it stretches to a distance x1. Hooke's Law states that:
[tex]F=k.x[/tex]
Where x is the elongation of the spring:
[tex]x=x1-x0[/tex]
We are given the characteristics of a spring of x0=31 cm. When a mass of m=8 Kg is hung from the spring, it stretches to x1=36.9 cm. We need to calculate the force of the mass of 8 Kg. It can be done by calculating the weight:
[tex]F = m.g=8\ Kg\cdot 9.8\ m/s^2[/tex]
[tex]F=78.4\ N[/tex]
The elongation of the spring is
[tex]x=36.9\ cm - 31\ cm = 5.9\ cm[/tex]
Converting to meters:
[tex]x=5.9/100=0.059\ m[/tex]
(a)
From Hooke's Law, we solve for k:
[tex]\displaystyle k=\frac{F}{x}=\frac{78.4}{0.059}[/tex]
[tex]k=1,329\ N/m[/tex]
(b) With the value of k, the equation for the spring is:
[tex]F=1,329.x[/tex]
Now if a weight of F=185 N is hung from the spring, the elongation is:
[tex]\displaystyle x=\frac{F}{1,329}=\frac{185}{1,329}[/tex]
[tex]x=0.1392\ m=13.92\ cm[/tex]
Thus, the length of the spring is:
[tex]x1=xo+x=31\ cm+13.92\ cm=44.92\ cm[/tex]
The length of the spring is 44.92 cm
