A spring extends 10 cm whena mass of 100g is attached to it, when is the spring constant
0.1 n/m
To determine the spring constant, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension or compression.
Hooke's Law can be expressed as:
F = -kx
Where:
F is the force exerted by the spring,
k is the spring constant, and
x is the displacement or extension of the spring.
In this case, we know that the spring extends 10 cm (which is equal to 0.1 meters) when a mass of 100 g (0.1 kg) is attached to it. So, x = 0.1 meters and F = weight of the mass.
The weight of the mass can be calculated using the formula:
Weight = mass × acceleration due to gravity
Considering the acceleration due to gravity as 9.8 m/s², the weight of the mass is:
Weight = 0.1 kg × 9.8 m/s²
Weight = 0.98 N
Now, substituting the known values into Hooke's Law:
0.98 N = -k × 0.1 m
Now, solving for k:
k = -0.98 N / 0.1 m
k = -9.8 N/m
Therefore, the spring constant is -9.8 N/m. The negative sign indicates that the spring is in the stretched position.
a hanging spring with a freely hanging mass
the force (from gravity) is
... m * g = .98 N
k = F / d = .98 N / .01 m = 98 N/m