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.