Use Hooke's Law for springs, which states that the distance a spring is stretched (or compressed) varies directly as the force on the spring.
A force of 265 newtons stretches a spring 0.15 meter (see figure).
(a) How far will a force of 120 newtons stretch the spring? (Round your answer to two decimal places.)
:m
(b) What force is required to stretch the spring 0.3 meter? (Round your answer to two decimal places.)
:N
a. d = (120N/265N) * 0.15m
b. (F/265) * 0.15 = 0.30
0.15F = 79.5
F = 530 N.
To solve both parts of this problem, we can use the formula for Hooke's Law, which states that the distance a spring is stretched (or compressed) varies directly as the force on the spring. The formula is written as:
F = kx
Where:
F = force applied to the spring
k = spring constant (a measure of the stiffness of the spring)
x = distance the spring is stretched or compressed
To solve Part (a), we are given the force (F = 265 N) and the distance the spring is stretched (x = 0.15 m). We need to find out how far a force of 120 newtons will stretch the spring.
We can set up a proportion based on Hooke's Law:
F1 / x1 = F2 / x2
Where F1 and x1 are the given values, and F2 and x2 are the unknown values we need to find.
Plugging in the values, we get:
265 N / 0.15 m = 120 N / x2
To solve for x2, we can cross-multiply and divide:
x2 = (120 N * 0.15 m) / 265 N
Calculating this, we get x2 ≈ 0.068 m. Therefore, a force of 120 newtons will stretch the spring approximately 0.068 meters.
To solve Part (b), we are given the distance the spring is stretched (x = 0.3 m). We need to find out what force is required to stretch the spring this distance.
Again, we can use Hooke's Law and set up a proportion:
F1 / x1 = F2 / x2
Plugging in the values, we get:
265 N / 0.15 m = F2 / 0.3 m
To solve for F2, we can cross-multiply and divide:
F2 = (265 N * 0.3 m) / 0.15 m
Calculating this, we get F2 ≈ 530 N. Therefore, a force of 530 newtons is required to stretch the spring 0.3 meters.