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.
(a) What force is required to stretch the spring 0.2 meter? (Round your answer to two decimal places.)
Incorrect: Your answer is incorrect. .
N
(b) How far will a force of 100 newtons stretch the spring? (Round your answer to two decimal places.)
m
since d/F is constant,
.15/265 = .2/x
and,
.165/265 = x/100
Note that you don't really have to find k, the constant of proportionality. You can go ahead and find it, but your answers will be the same.
To use Hooke's Law for springs, we need to find the proportionality constant known as the spring constant (k). This constant represents the stiffness or rigidity of the spring and is unique to each spring.
Hooke's Law can be written as follows:
F = k * x,
where F is the force applied to the spring, x is the distance the spring is stretched (or compressed), and k is the spring constant.
To find the spring constant (k), we can rearrange the equation as follows:
k = F / x,
where k is the spring constant, F is the force applied, and x is the distance the spring is stretched.
(a) To find the force required to stretch the spring 0.2 meters, we can use Hooke's Law and the given force and distance:
k = 265 N / 0.15 m,
k = 1766.67 N/m (rounded to two decimal places).
Now, we can use the spring constant (k) to find the force:
F = k * x,
F = 1766.67 N/m * 0.2 m,
F = 353.33 N (rounded to two decimal places).
Therefore, the force required to stretch the spring 0.2 meters is approximately 353.33 N.
(b) To find how far a force of 100 Newtons will stretch the spring, we can rearrange Hooke's Law to solve for x:
x = F / k,
where x is the distance the spring is stretched, F is the force applied, and k is the spring constant.
Using the spring constant (k) we found in part (a):
x = 100 N / 1766.67 N/m,
x = 0.0566 m (rounded to two decimal places).
Therefore, a force of 100 Newtons will stretch the spring by approximately 0.0566 meters.