A coil spring stretches 4 cm when a mass of 500 g is suspended from it . what is the force constant of the spring.
If this is on Earth, then the force of gravity is .5kg*9.8N/kg=4.9Newtons
forceconstant=force/stretchinMeters
=4.9/.04 N/m
The force constant (k) of a spring describes how stiff it is and can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Mathematically, Hooke's Law can be written as:
F = k * x
Where:
F is the force applied to the spring,
k is the force constant of the spring, and
x is the displacement from the equilibrium position.
In this case, the given information states that a coil spring stretches 4 cm (0.04 m) when a mass of 500 g (0.5 kg) is suspended from it.
Using Hooke's Law, we can write:
0.5 kg * 9.8 m/s² = k * 0.04 m
Simplifying the equation:
4.9 N = 0.04 k
To find k, we divide both sides of the equation by 0.04:
k = 4.9 N / 0.04 m = 122.5 N/m
Therefore, the force constant of the spring is 122.5 N/m.
To find the force constant of the spring, we need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
Hooke's Law can be expressed as:
F = k * x
Where:
F is the force applied to the spring
k is the force constant (also known as the spring constant)
x is the displacement from the equilibrium position
In this case, we know that when a mass of 500 g (0.5 kg) is suspended from the spring, the spring stretches by 4 cm (0.04 m).
So, the force applied to the spring (F) is equal to the weight of the mass, which can be calculated using:
F = m * g
Where:
m is the mass (0.5 kg)
g is the acceleration due to gravity (approximately 9.8 m/s^2)
F = 0.5 kg * 9.8 m/s^2
F = 4.9 N
Now, we can rearrange Hooke's Law and solve for the force constant (k):
k = F / x
k = 4.9 N / 0.04 m
k = 122.5 N/m
Therefore, the force constant (or spring constant) of the coil spring is 122.5 N/m.