Kati observes a force of 15N stretch a spring a distance of 0.450m. A) What is the elasticity coefficient of the spring and B) how much force must be done to stretch the spring a distance of 0.750m?
a) F=kx
k=33.3N ?
b)???
For the second part use the same equation
Sub k (what you found earlier) and x=0.750 to get F
F=24.9N?
and I dividing the f/x or do I multiply
A. k = 15N/0.450m = 33.3 N/m.
B. F = 0.75m * 33.3N/m = 25 N.
To find the elasticity coefficient of the spring, we can use Hooke's Law equation, which states that the force required to stretch or compress a spring is proportional to the displacement.
Hooke's Law equation: F = kx
Where:
F is the force applied to the spring
k is the elasticity coefficient or spring constant
x is the displacement or stretch/compression distance
Given values:
F = 15N
x = 0.450m
a) To find the elasticity coefficient (k):
Step 1: Rearrange the equation to solve for k:
k = F/x
Step 2: Substitute the given values into the equation:
k = 15N / 0.450m
k ≈ 33.33 N/m
So, the elasticity coefficient of the spring is approximately 33.33 N/m.
b) Now, to find the force required to stretch the spring a distance of 0.750m:
Step 1: We will use the same Hooke's Law equation: F = kx
Step 2: Substitute the known values into the equation:
F = (33.33 N/m) * (0.750m)
F ≈ 25 N
Therefore, the force required to stretch the spring a distance of 0.750m is approximately 25N.