A spring has a length of 0.190 m when a 0.300 kg mass hangs from it, and a length of 0.750 m when a 2.80 kg mass hangs from it.
(a) What is the force constant of the spring?
(b) What is the unloaded length(m) of the spring?
x is the length of unloaded spring
F=k•Δx
F=mg =>
m₁g=k(0.19-x)
m₂g=k(0.75-x)
m₁/m₂=(0.19-x)/(0.75-x)
x=[0.19m₂ - 0.75m₁]/(m₂ -m₁) =
=[0.19•2.8 – 0.75•0.3]/(2.8-0.3) = 0.1228 m
k= m₁g/(0.19-x) = 0.3•9.8/(0.19-0.1228)= 43.75 N/m
To find the force constant of the spring and the unloaded length of the spring, we can use Hooke's Law. Hooke's Law states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position. Mathematically, it is expressed as F = -kx, where F is the force, k is the force constant of the spring, and x is the displacement.
Let's solve the problem step by step:
(a) To find the force constant of the spring:
The force applied to the spring is the weight of the mass hanging from it. We can calculate the force using the formula F = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s^2).
For the first scenario:
Mass (m1) = 0.300 kg
Force (F1) = m1 * g = 0.300 kg * 9.8 m/s^2 = 2.94 N
For the second scenario:
Mass (m2) = 2.80 kg
Force (F2) = m2 * g = 2.80 kg * 9.8 m/s^2 = 27.44 N
Next, we can calculate the displacement of the spring in each scenario. The displacement is the difference between the spring's initial length (unloaded length) and the stretched or compressed length.
For the first scenario:
Displacement (x1) = initial length - stretched length = 0.190 m - 0 = 0.190 m
For the second scenario:
Displacement (x2) = initial length - stretched length = 0.750 m - 0 = 0.750 m
Now, using Hooke's Law, F1 = -k * x1 and F2 = -k * x2.
Substituting the values, we have:
2.94 N = -k * 0.190 m
27.44 N = -k * 0.750 m
To find the force constant (k), we can rearrange the equation:
k = -F / x
For the first scenario:
k1 = -2.94 N / 0.190 m = -15.47 N/m (rounded to two decimal places)
For the second scenario:
k2 = -27.44 N / 0.750 m = -36.59 N/m (rounded to two decimal places)
Since the force constant of the spring should be the same regardless of the mass, we can take the average of the two values:
k_avg = (k1 + k2) / 2 = (-15.47 N/m + (-36.59 N/m)) / 2 = -26.03 N/m (rounded to two decimal places)
Therefore, the force constant of the spring is approximately -26.03 N/m.
(b) To find the unloaded length (equilibrium position) of the spring:
Since the equilibrium position is when there is no net force acting on the spring, we know that the force applied to the spring (Funloaded) is zero.
Using Hooke's Law, we have:
Funloaded = -k * xunloaded
We can rearrange the equation to solve for the unloaded length:
F = -k * xunloaded
0 = -k * xunloaded
k * xunloaded = 0
Since k is a non-zero value (as we found earlier), for the equation to hold, xunloaded must be zero.
Therefore, the unloaded length of the spring is 0 meters.
So, the answer for part (a) is approximately -26.03 N/m and for part (b), the unloaded length of the spring is 0 meters.