A vertical spring has a length of 0.175m when a 0.25kg mass hangs from it, and a length of 0.775m when a 1.9kg mass hangs from it. What is the force constant of the spring in N/m?

[g * (1.9 - .25)] / (.775 - .175)

5.95

To find the force constant of the spring, we can 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 is expressed as: F = -kx

Where:
F = force applied to the spring (in Newtons)
k = force constant (in N/m)
x = displacement from the equilibrium position (in meters)

We are given two cases:
Case 1: When a 0.25kg mass hangs from the spring, the length of the spring is 0.175m.
Case 2: When a 1.9kg mass hangs from the spring, the length of the spring is 0.775m.

Let's calculate the displacement (x) for each case:

Displacement (x1) in Case 1:
x1 = length of the spring - equilibrium length
= 0.175m - 0m (equilibrium length)
= 0.175m

Displacement (x2) in Case 2:
x2 = length of the spring - equilibrium length
= 0.775m - 0m (equilibrium length)
= 0.775m

Next, we can use Hooke's Law to determine the force constant (k) for each case.

Case 1:
F1 = -k * x1
Since the force applied is equal to the weight of the mass (F = mg):
F1 = -0.25kg * 9.8m/s^2 (acceleration due to gravity)
F1 = -2.45N

Case 2:
F2 = -k * x2
F2 = -1.9kg * 9.8m/s^2
F2 = -18.62N

Now, we can equate the two expressions for force (F1 and F2) and solve for k:

-2.45N = -18.62N * 0.775m / 0.175m

Simplifying, we have:

2.45N / 18.62N = 0.775m / 0.175m

0.1319 = 4.4286

Therefore, the force constant (k) of the spring is approximately:

k = F1 / x1
k = -2.45N / 0.175m
k = -14N/m

Please note that the negative sign indicates that the force exerted by the spring acts in the opposite direction of the displacement. In this case, the negative sign implies that the force exerted by the spring is directed upward.

To find the force constant of the spring, 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.

The formula for Hooke's law is:

F = k * x

Where:
F is the force exerted by the spring,
k is the force constant (also known as the spring constant),
x is the displacement of the spring from its equilibrium position.

We are given the lengths of the spring with different masses attached. The displacement of the spring is the difference in lengths:

Δx = x2 - x1

First, let's calculate the displacement Δx:

Δx = 0.775 m - 0.175 m
= 0.6 m

Next, we need to find the corresponding forces for the two masses. We can use the formula for gravitational force:

F = m * g

Where:
F is the force,
m is the mass,
g is the acceleration due to gravity (approximately 9.8 m/s^2).

For the 0.25 kg mass:

F1 = 0.25 kg * 9.8 m/s^2
= 2.45 N

For the 1.9 kg mass:

F2 = 1.9 kg * 9.8 m/s^2
= 18.62 N

Now, we can use Hooke's law to find the force constant:

F = k * Δx

Rearranging the formula, we get:

k = F / Δx

For the 0.25 kg mass:

k1 = 2.45 N / 0.6 m
= 4.08 N/m (rounded to two decimal places)

For the 1.9 kg mass:

k2 = 18.62 N / 0.6 m
= 31.03 N/m (rounded to two decimal places)

Since we have two values for the force constant, we can take the average:

k_avg = (k1 + k2) / 2

k_avg = (4.08 N/m + 31.03 N/m) / 2
= 17.55 N/m (rounded to two decimal places)

Therefore, the force constant of the spring is approximately 17.55 N/m.