Three different masses are attached to a spring which hangs down vertically from the ceiling. A metric ruler hangs from the ceiling next to the spring and measures how far from the ceiling the sping hangs. When a 110 N weight is attached to the spring, the spring reaches the 40 mm mark on the ruler. When a 240 N weight is attached to the sping, it hangs down to the 60 mm mark. An unknown weight is attached to the sping and it hangs down 30 mm.
a) How far down will the spring hang when there is no weight on the spring
Are you in neils class too?
yess! I can't figure this one out.
To find how far down the spring will hang when there is no weight on it, we can use the concept of Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position.
Let's denote the displacement of the spring from its equilibrium position as x (in mm), and the force exerted by the spring as F (in N). We can assume that the spring obeys Hooke's Law, which can be expressed as:
F = k · x
where k is the spring constant. In our case, we have two data points: when a 110 N weight is attached, the displacement is 40 mm, and when a 240 N weight is attached, the displacement is 60 mm.
Using these two data points, we can calculate the spring constant (k) by rearranging Hooke's Law:
k = F / x
For the first data point:
k = 110 N / 40 mm = 2.75 N/mm
For the second data point:
k = 240 N / 60 mm = 4 N/mm
Now that we have calculated the spring constant, we can determine how far down the spring will hang when there is no weight on it. Let's denote this unknown weight as W (in N) and the displacement as x (in mm).
Using Hooke's Law:
F = k · x
W = k · x
Rearranging this equation for x:
x = W / k
Plugging in the known values:
x = 30 mm / (4 N/mm) = 7.5 mm
Therefore, the spring will hang down 7.5 mm when there is no weight on it.