a force of 10n causes a spring to extend by 20mm and what is sprinconstant
F = kx
so plug in your numbers
To find the spring constant, we can use Hooke's Law, which states that the force applied to a spring is directly proportional to the extension or compression of the spring.
Hooke's Law can be written as:
F = k * x
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement or extension of the spring
In this case, the force applied (F) is 10 N and the extension (x) is 20 mm (or 0.020 m).
Using the Hooke's Law equation, we can rearrange it to solve for the spring constant (k):
k = F / x
Substituting the given values:
k = 10 N / 0.020 m
Simplifying:
k = 500 N/m
Therefore, the spring constant is 500 N/m.
To find the spring constant, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the amount the spring is stretched or compressed.
Hooke's Law can be expressed as:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant, and
x is the displacement (stretch or compression) of the spring.
We have the force (F) as 10 N and the displacement (x) as 20 mm (which needs to be converted to meters).
First, we convert 20 mm to meters:
1 meter = 1000 millimeters
So, 20 mm = 20 / 1000 = 0.02 meters
Now, we can rearrange Hooke's Law equation to solve for the spring constant (k):
k = F / x
Substituting the given values:
k = 10 N / 0.02 m
k = 500 N/m
Therefore, the spring constant is 500 N/m.