A spring is stretched by 8 cm when a force of 2 N is applied. If a force of 5.8 N is applied, how many additional cm does the spring stretch?
would it be 8/2 =4*5.8= 23.2
yup
To calculate the additional stretching of the spring, you need to use Hooke's Law, which states that the force applied to a spring is proportional to the displacement or stretching of the spring.
First, let's calculate the spring constant (k) using the initial stretching and force applied:
k = F / x
where k is the spring constant, F is the force applied, and x is the stretching of the spring.
In this case:
F = 2 N (initial force)
x = 8 cm (initial stretching)
So, k = 2 N / 8 cm = 0.25 N/cm
Now, you can use this spring constant to calculate the additional stretching of the spring when a force of 5.8 N is applied:
x2 = F2 / k
where x2 is the additional stretching of the spring, F2 is the additional force applied, and k is the spring constant (0.25 N/cm).
In this case:
F2 = 5.8 N (additional force)
x2 = 5.8 N / 0.25 N/cm = 23.2 cm
Therefore, when a force of 5.8 N is applied, the spring will stretch an additional 23.2 cm.