What is the restoring force of a spring stretched 0.35 meters with a spring contact of 55 newtons/meter?
F = 55N/m * 0.35m = 19.25 N.
The restoring force of a spring can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be expressed as: F = -kx
Where:
F is the restoring force applied by the spring,
k is the spring constant (also known as the spring contact or stiffness constant),
x is the displacement of the spring from its equilibrium position.
In this case, we are given the displacement x as 0.35 meters and the spring constant k as 55 newtons/meter.
To calculate the restoring force, we can substitute the given values into the formula:
F = -kx
F = -55 N/m * 0.35 m
Now, let's calculate the restoring force:
F = -55 N/m * 0.35 m
F = -19.25 Newtons
Therefore, the restoring force of the spring when stretched 0.35 meters is -19.25 Newtons. The negative sign indicates that the force is acting in the opposite direction of the displacement, which is towards the equilibrium position (opposite to the stretching direction).