In slowly stretching a certain spring i pull harder and...

In slowly stretching a certain spring, I pull harder and harder until it is stretched by a distance of 2 = 0.5 m. The force required to hold it at this point is F = 10 N. How much work did I do in stretching it out to a distance of 0.5 m?

Transcript

0:00 Hi there.

00:01 So for this problem, we are told that it takes 100 joules of work.

00:06 So the work done is 100 joules to stretch this spring to a distance of 0 .5 meters, and this is from the equilibrium position.

00:18 So the question is, how much work is needed to stretch it at an additional 0 .5 meters? so what we need to do is with the ballast that we are given, we are going to obtain the spring's constant.

00:29 Okay.

00:30 So we know first that the word done in order to stretch this to 0 .5 meters from the natural equilibrium.

00:41 It's just one divided by two times the spring's concept times this value delta s.

00:46 Now we're going to solve for the value of k.

00:49 Oh sorry, it's delta s to the square.

00:52 So let's solve for the concept of proportionality.

00:55 So the concept of proportionality is equal to two times the word done, divided by delta x to the square.

01:03 So now we just need to simply substitute the value.

01:05 So that will be 2 times 200.

01:08 And this divided by 0 .5 and that to the square.

01:11 So let's see what value for the concept of proportionality.

01:15 We obtained from this.

01:21 Okay, so let's see.

01:26 So in this case, we obtain 1 ,600.

01:31 And this in units of newtoms per meter.

01:35 Now, once we have this, we use the same expression as before, but in this case, this is the spring's comes and the distance ends.

01:43 Now, we go from the 0 .5 because we started that because we are told to an additional 0 .5.

01:50 So we go from 0 .5 to an additional 0 .5.

01:54 So that will be 0 .5 plus 0 .5, which is equal to 1.

01:58 And now we just evaluate this...

Question

In slowly stretching a certain spring, I pull harder and harder until it is stretched by a distance of 2 = 0.5 m. The force required to hold it at this point is F = 10 N. How much work did I do in stretching it out to a distance of 0.5 m?

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