A force of 10N causes a spring to extend by 20 mm find the extension of the spring when 25 N is.applied
To find the extension of the spring when a force of 25 N is applied, we can use Hooke's Law, which states that the force required to stretch or compress a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be written as:
F = k * x,
where F is the force applied to the spring, k is the spring constant, and x is the displacement or extension of the spring.
In this case, we have a force of 10 N that causes a spring to extend by 20 mm. Let's use this information to find the spring constant k.
10 N = k * 20 mm.
First, we need to convert millimeters to meters for consistency.
20 mm = 20 / 1000 = 0.02 m.
Now, we can calculate k:
k = 10 N / 0.02 m = 500 N/m.
We have determined that the spring constant is 500 N/m.
Now, we can find the extension of the spring when a force of 25 N is applied.
25 N = 500 N/m * x.
Solving for x:
x = 25 N / 500 N/m = 0.05 m.
Finally, let's convert the extension to millimeters:
0.05 m = 0.05 * 1000 = 50 mm.
Therefore, the extension of the spring when a force of 25 N is applied is 50 mm.