What is the spring constant of Spring 1, in N/m?
This is in an online lab. All I know about spring 1 is that it is 30cm long, and with a mass of 250g attached, it reaches 55cm long.
My answer of 4.5 (.25 * 10 = k * .55) is incorrect...why? Thank you!
Got it - I wasn't taking the magnitude of displacement :)
Newtons is force ... like m g
m g = k x
.25 kg * 9.8 = k * .25 m
m g = F = k x = k(.55-.30)
.25(9.81) = k (.25)
k = g = 9.81 N/m
To determine the spring constant (k) of Spring 1, you can use Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. The formula is given as:
F = kx
Where:
F is the force applied on the spring (in Newtons),
k is the spring constant (in N/m),
x is the displacement of the spring from its equilibrium position (in meters).
Based on the information provided, the initial length of the spring is 30 cm (or 0.3 m), and when a mass of 250 g (or 0.25 kg) is attached, the spring reaches a length of 55 cm (or 0.55 m).
To calculate the spring constant, we need to determine the displacement (x) and the force (F).
The displacement (x) can be calculated as the difference between the final length and the initial length:
x = 0.55 m - 0.30 m
x = 0.25 m
Next, we need to calculate the force (F) using the mass (m) and the acceleration due to gravity (g):
F = m * g
where the acceleration due to gravity (g) is approximately 9.8 m/s^2.
F = 0.25 kg * 9.8 m/s^2
F = 2.45 N
Now, we can rearrange Hooke's Law to solve for the spring constant (k):
k = F / x
k = 2.45 N / 0.25 m
k = 9.8 N/m
Therefore, the spring constant of Spring 1 is 9.8 N/m, not 4.5 N/m as you calculated. It seems there was an error in calculating the force or the displacement. Double-check your calculations to ensure accurate results.