a) The spring constant is 1225 N/m
b) The mass of the fish is 6.88 kg
c) The marks are 0.4 cm apart
Explanation:
a)
When the spring is at equilibrium, the weight of the load applied to the spring is equal (in magnitude) to the restoring force of the spring, so we can write
[tex]mg = kx[/tex]
where
m is the mass of the load
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
k is the spring constant
x is the stretching of the spring
For the load in this problem we have
m = 10.0 kg
x = 8.00 cm = 0.08 m
Substituting, we find the spring constant
[tex]k=\frac{mg}{x}=\frac{(10)(9.8)}{0.08}=1225 N/m[/tex]
b)
As before, at equilibrium, the weight of the fish must balance the restoring force in the spring, so we have
[tex]mg=kx[/tex]
where this time we have:
m = mass of the fish
[tex]g=9.8 m/s^2[/tex]
k = 1225 N/m is the spring constant
x = 5.50 cm = 0.055 m is the stretching of the spring
Substituting,
[tex]m=\frac{kx}{g}=\frac{(1225)(0.055)}{9.8}=6.88 kg[/tex]
c)
To solve this part, we just need to find the change in stretching of the spring when a load of half-kilogram is hanging on the spring. Using again the same equation,
[tex]mg=kx[/tex]
where this time we have:
m = 0.5 kg
[tex]g=9.8 m/s^2[/tex]
k = 1225 N/m
x = ? is the distance between the half-kilogram marks on the scale
Substituting,
[tex]x=\frac{mg}{k}=\frac{(0.5)(9.8)}{1225}=0.004 m = 0.4 cm[/tex]
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