Karen hang a spring vertically and attaches 30g Mass to the springs other end. If the spring stretches 5 cm. What is the spring constant?
F =-k x
-m g = -k x
(0.030)(9.81) = k (0.05)
k = (0.030)(9.81)/(0.05) Newtons/meter
To calculate the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be written as:
F = kx
Where:
F is the force applied on the spring (in newtons),
k is the spring constant (in newtons per meter), and
x is the displacement of the spring from its equilibrium position (in meters).
In this case, we know that the spring stretches 5 cm (which is equivalent to 0.05 meters) when a 30g mass is attached. To find the spring constant, we need to calculate the force applied to the spring.
The force applied can be calculated using the formula:
F = mg
Where:
m is the mass (in kilograms), and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
First, we need to convert the mass from grams to kilograms:
m = 30g = 30/1000 kg = 0.03 kg
Now, we can calculate the force applied:
F = mg = (0.03 kg) * (9.8 m/s^2) = 0.294 N
Finally, we can rearrange Hooke's Law to solve for the spring constant:
k = F/x
k = 0.294 N / 0.05 m
k = 5.88 N/m
Therefore, the spring constant is 5.88 N/m.