Answer: 94 cm
Explanation:
The answer is not among the given options, however it can be approximated to 94 cm.
Hooke's law establishes that the elongation of a spring is directly proportional to the modulus of the force [tex]F[/tex] applied to it, as long as the spring is not permanently deformed:
[tex]F=k (x-x_{o})[/tex] (1)
Where:
[tex]k[/tex] is the elastic constant of the spring
[tex]x_{o}[/tex] is the length of the spring without applying force.
[tex]x[/tex] is the length of the spring with the force applied.
According to this, we have a spring where only the force due gravity is applied. In other words, the force applied is the weigth [tex]W[/tex]:
[tex]W=m.g[/tex] (2)
So, equation (1) can be rewritten as:
[tex]W=k (\Delta x)[/tex] (3)
Isolating [tex]k[/tex]:
[tex]k=\frac{W}{\Delta x}[/tex] (4) From here we can deduce [tex]k[/tex] is the slope of the line of the graph shown, as follows:
[tex]k=\frac{m_{2}g-m_{1}g}{x-x_{o}}=g\frac{m_{2}-m_{1}}{x-x_{o}}[/tex]
Where:
[tex]m_{2}=400 g=0.4 kg[/tex]
[tex]m_{1}=200 g=0.2 kg[/tex]
[tex]g=9.8\frac{m}{s^{2}}[/tex] is the acceleration due gravity
[tex]x=15 cm=0.15 m[/tex]
[tex]x_{o}=7.5 cm=0.075 m[/tex]
[tex]k=9.8\frac{m}{s^{2}}\frac{0.4 kg-0.2 kg}{0.15 m-0.075 m}[/tex]
[tex]k=26.13 N/m[/tex] (5) This is the spring constant
Now, with the mass [tex]m=2.35 kg[/tex] we can calculate the weight:
[tex]W=m.g=(2.35 kg)(9.8\frac{m}{s^{2}})[/tex]
[tex]W=23.03 N[/tex] (6)
So, knowing [tex]k=26.13 N/m[/tex] and [tex]W=23.03 N[/tex] we can calculate [tex]\Delta x[/tex] from (3):
[tex]\Delta x=\frac{W}{k}[/tex] (7)
[tex]\Delta x=\frac{23.03 N}{26.13 N/m}[/tex] (8)
[tex]\Delta x=0.88 m=88.13 cm[/tex] (9)
The value that is nearest to 88.13 cm is 94 cm
