a spring with a length of 22.0 centimeters hangs from a ceiling. when a 755-gram mass is attached ti the spring, the spring strethes to 35.0 centimeters. what is the magnitude of the spring constant in N/m
The stretching distance is 35 -22 = 13 cm
= 0.13 m
The force is M*g = 0.755kg*9.81 m/s^2
= 7.41 N
The spring constant is 7.41/0.13 = ___ N/m
To find the magnitude of the spring constant in N/m, we need to use Hooke's law, which states that the force exerted by a spring is directly proportional to its displacement.
The formula for Hooke's law is F = k * x, where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.
In this case, we know the following information:
- The length of the relaxed spring (equilibrium position) is 22.0 centimeters.
- When a 755-gram mass is attached to the spring, it stretches to a length of 35.0 centimeters.
To find the displacement (x), we subtract the equilibrium position from the stretched position:
x = 35.0 cm - 22.0 cm = 13.0 cm
Now, we need to convert the masses and lengths to SI units:
- Convert the mass from grams to kilograms:
755 grams = 0.755 kg
- Convert the lengths from centimeters to meters:
22.0 cm = 0.22 m
35.0 cm = 0.35 m
13.0 cm = 0.13 m
Now we can plug these values into Hooke's law and solve for the spring constant (k):
F = k * x
0.755 kg * 9.8 m/s^2 = k * 0.13 m
Rearranging the equation, we get:
k = (0.755 kg * 9.8 m/s^2) / 0.13 m
Evaluating the equation, we get:
k = 56.46 N/m
Therefore, the magnitude of the spring constant is 56.46 N/m.