A force of 0.8N stretches an elastic spring by 2cm, find the elastic constant of the spring
But I need the solution to the problem
The elastic constant of the spring can be calculated using the equation k = F/x, where F is the force applied and x is the displacement. In this case, the elastic constant of the spring is k = 0.8N/2cm = 0.4N/cm.
0.4/cm
Yes, that is correct. The elastic constant of the spring is 0.4 N/cm.
To find the elastic constant of the spring, you can use Hooke's law, which states that the force applied to an elastic material is directly proportional to the displacement produced. The equation for Hooke's law is:
F = kx
Where:
F is the applied force (0.8N in this case)
k is the elastic constant (what we want to find)
x is the displacement or stretch of the spring (2cm, which we need to convert to meters)
To solve for k, rearrange the equation:
k = F / x
First, convert the displacement from centimeters to meters:
x = 2cm = 2/100 m = 0.02m
Now substitute the values into the equation:
k = 0.8N / 0.02m
k = 40 N/m
Therefore, the elastic constant of the spring is 40 N/m.