The amount of force required to compress a spring is inversely proportional to the distance that it has been compressed. For one certain spring, it takes 5 pounds of force to compress it from its natural length of 18 inches down to 12 inches. How much force would be required to compress the spring down to 9 inches? Thank you. Is there a formula for this not sure how to approach it...

F = k*deltax

F=force
k= spring constant
deltax= change in distance

find spring constant k:

5lb = k*(12-18)in

k = 5lb / (12-18)in

k= -0.8333 lb/in

force to compress spring to 9inches:

F = -0.8333lb/in * (9-18)in

Sorry, I didn't solve it in a calculus manner.

F= ∫kdx with upper and lower limits of distance

Integrating:

F= k*(x_f - x_i)

x_f = final distance , x_i = initial distance

and the rest should be the same as above post.

Yes, there is a formula that relates the force required to compress a spring to the distance it has been compressed. In this case, you are given that the force and distance are inversely proportional. The formula for an inverse proportion can be written as:

Force = constant / distance

In this case, the constant can be found by using the initial force and distance given. Let me help you calculate it step by step:

1. Start by finding the constant. Given that it takes 5 pounds of force to compress the spring from 18 inches to 12 inches, we can set up the equation:

5 = constant / 6

Now we can solve for the constant:

constant = 5 * 6 = 30

So, the constant in this case is 30.

2. Now that we have the constant, we can use it to find the force required to compress the spring to 9 inches. Using the same formula:

Force = constant / distance

Force = 30 / 9 = 3.33 pounds (approximately)

Therefore, approximately 3.33 pounds of force would be required to compress the spring down to 9 inches.

Remember, the formula to determine the force required to compress a spring is:

Force = constant / distance

You can use this formula whenever you have the constant and the distance to find the force, or vice versa.