A spiral spring extend from a length of 10.01cm to 10.01cm when a force of 20n is applied on it. calculate the force of the spring
You have two typos that make this impossible
10.01 to 10.01 is no stretch at all. When you get it right remember to convert to meters.
You said the force was 20 Newtons
I suspect you want the force constant k in F = k x
where x is the stretch in meters.
If so k = 20 N / stretch in meters
To calculate the force of the spring, you need to use Hooke's Law, which states that the force exerted by a spring is directly proportional to the extension or compression of the spring.
Hooke's Law is expressed as F = -kx, where F is the force applied on the spring, k is the spring constant, and x is the extension or compression of the spring from its equilibrium position.
In this case, the spring extends by a length of 10.01 cm from its equilibrium position when a force of 20 N is applied. To use Hooke's Law, we need to convert the extension to meters:
10.01 cm = 0.1001 m
Now, we can rearrange Hooke's Law to solve for the force (F):
F = -kx
Solving for k:
k = -F / x
Substituting the given values:
k = -20 N / 0.1001 m = -199.8002 N/m
Now we can use the spring constant (k) to find the force. In this case, we need to find the force when the spring is not extended or compressed, i.e., at its equilibrium position where the length is 10.01 cm.
To find the force at the equilibrium position, we use the equation:
F = -kx
Substituting the values:
F = -(-199.8002 N/m) * (0 m) = 0 N
Therefore, the force of the spring at its equilibrium position is 0 N.