If the length of a #48# #cm# spring increases to #93# #cm# when a #6# #kg# weight is hanging from it, what is the spring's constant?

Question

Olivia Askew · Accepted Answer

First, we need to work in SI units - meters, not centimeters.

The length increases from #0.48# #m# to #0.93# #m#. The force acting is the weight force, #F=mg=6*9.8=58.8# #N#.

Then #k=F/x=58.8/(0.93-0.48)~~130.7# #Nm^-1# (or #kgs^-2#).

Olivia Askew · Answer

undefined The spring constant can be calculated using 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 to calculate the spring constant (k) is:

k = (F) / (x)

Where:
- k is the spring constant,
- F is the force exerted on the spring,
- x is the displacement from the equilibrium position.

In this case, the displacement (x) is the change in length of the spring, which is 93 cm - 48 cm = 45 cm = 0.45 m. The force exerted by the weight (F) can be calculated using the formula:

F = m * g

Where:
- m is the mass (6 kg),
- g is the acceleration due to gravity (approximately 9.8 m/s^2).

Substituting the values into the equation:

F = 6 kg * 9.8 m/s^2 = 58.8 N

Now, plug the values into the formula for the spring constant:

k = (58.8 N) / (0.45 m) ≈ 130.67 N/m

So, the spring constant is approximately 130.67 N/m.