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
where
m is the mass of the load
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
b)
As before, at equilibrium, the weight of the fish must balance the restoring force in the spring, so we have
where this time we have:
m = mass of the fish
k = 1225 N/m is the spring constant
x = 5.50 cm = 0.055 m is the stretching of the spring
Substituting,
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,
where this time we have:
m = 0.5 kg
k = 1225 N/m
x = ? is the distance between the half-kilogram marks on the scale
Substituting,
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