# Thread: Acceleration and Deceleration - The breaking point of a rope

1. Hello guys,

I need some help finding the right formula for calculating the the amount of extra kilograms that is put on an object when it's dropped from a certain height and then stopped. Because i want to see how much a paracord can take. I know that a 550 paracord can withstand a static weight of 550 lb. But what if I dropped a 45 lb object from 6 ft? How much weight did the paracord have to stop?

I am sorry for any spelling mistakes. English is not my mothertongue and neither is physics, hence my question , Demco  2.

3. The answer will depend on the length and stretchiness of the parachute cord.
The best way to work it out is to compare the gravitational potential energy of the falling weight with the elastic potential energy the cord absorbs before it breaks.

Let's say your cord is 6 feet long and stretches 30% before it breaks. This means that the cord will stretch 1.8 feet and will be 7.8 feet long at the breaking point. If the tension in the cord is zero at rest and 550 lb at maximum extension, the average tension throughout is (0+550)/2=275 lb, and the energy absorbed is 275*1.8=495 ft-lbf. The falling weigh of 45 lb will have 45lb*7.8ft=351 ft-lbf of potential energy. 495>351. Therefore there is not enough gravitational potential energy there to break the cord.  4. Hey Harold,

But I don't understand why it has to be divided by 2? Can you please explain that for me?

(0+550)/2=275 lb
The equations does make sense to me, I just want to fully understand it   5. When you average two numbers, you add them and divide by two. Before we start stretching the rope, the force is zero. At the end of the stretching, which is where it breaks, it is 550. The average of zero and 550 is 275.

Here is a discussion of the energy stored in a spring.
Potential Energy Stored in a Spring  550, acceleration, decelaration, paracord 