Answer:
(a) 1225 N/m
(b) 6.875 kg
(c) 0.4 cm
Explanation:
From hook's law,
F = ke ................. Equation 1
Where F = Force, k = spring constant, e = extension.
From the question,
Note: The force acting on the spring is the weight of the fish.
F = mg.
Where m = mass of the fish, g = acceleration due to gravity.
mg = ke.................. Equation 2
make k the subject of the equation,
k = mg/e............... Equation 3
Also make m the subject of the equation,
m = ke/g............. Equation 4
also make e the subject of the equation
e = mg/k............... Equation 5
(a)
using equation 3
k = mg/e
Given:
e = 8 cm = 0.08 cm, m = 10 kg, g = 9.8 m/s²
Substitute into equation 2
k = (10×9.8)/0.08
k = 1225 N/m.
(b)
If it stretches by 5.5 cm,
Given: e = 5.5 cm = 0.055 cm, k = 1225 N/m, g = 9.8 m/s²
Substitute into equation 4
m = (1225×0.055)/9.8
m = 6.875 kg.
(c)
Given: m = 0.5 kg, k = 1225 N/m, g = 9.8 m/s²
Substitute into equation 5
e = (0.5×9.8)/1225
e = 0.004 m
e = 0.4 cm.
(a) The force constant of the spring scale is 1225 N/m.
(b) The mass of the spring when it stretches 5.5 cm is 6.88 kg.
(c) The distance between half-kilogram marks on the scale is 0.4 cm.
The given parameters:
The spring constant is calculated by applying Hook's law as follows:
[tex]mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{10 \times 9.8}{0.08} \\\\k = 1225 \ N/m[/tex]
The mass of the spring when it stretches 5.5 cm;
[tex]mg = kx\\\\m = \frac{kx}{g} \\\\m = \frac{1225 \times 0.055}{9.8} \\\\m = 6.88 \ kg[/tex]
The distance between half-kilogram (0.5 kg);
[tex]mg = kx\\\\x = \frac{mg}{k} \\\\x = \frac{0.5 \times 9.8}{1225} \\\\x = 0.004 \ m\\\\x = 0.4 \ cm[/tex]
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