# Newton‘s 2nd Law: Example, 1 Dimension

by Jared Rovny

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00:01 Let's go ahead a try a one dimensional problem for Newton's second law.

00:05 Suppose I asked you if you accelerate a big 1.000 kilogram spaceship but you only applied one Newton of force, which by the way is approximately the weight of an apple.

00:14 Ironically enough, it's a very small force, and that object, the spaceship is starting from rest in space, starting from rest meaning it has no initial velocity.

00:23 We put it in space just so we know there's no gravity or anything else that we have to worry about, just the force on an object.

00:29 The question is, how far will the spaceship go in 30 days? So in about a month's time was a tiny amount of force acting on it.

00:37 So first pause, see if you can solve this, see if you can find out how far the spaceship has travelled in 30 days with a tiny, tiny force acting on it.

00:44 Now if you solve this problem, you can see an example of how to do so here.

00:49 So we're gonna run through this very quickly with an object moving under the influenced of a very tiny force of one Newton.

00:55 The first thing that we wanna do is recognize that if we're going to do any equation of motion type analysis, then we could write down an equations of motion.

01:04 For example, this one for the motion of an object under some acceleration, the problem of course in this problem is that we're not given an acceleration so if we look at this equations of motion, the distance, the final position minus the initial position which will be exactly this distance here where you see X and X0, is equal to the initial velocity times the time where the initial velocity is zero, so we can carry with that term plus one half the acceleration times the time square.

01:38 So we are already, almost done with this problem.

01:40 We need to put in the time which we know is 30 days.

01:43 We also need to put in the acceleration so we have to find that one still.

01:46 The time is not so bad.

01:48 We have 30 days and the first thing we have to do here is some unit analysis because we do want all of our units to be in our standard units, our kilograms, our meters and our seconds, and so we need to get days converted into seconds, so this is a great practice in unit analysis which I recommend you try doing first before you see me do it here.

02:09 To convert the units of days into seconds, we will keep multiplying by what our essential units of one.

02:16 In other words, there are 24 hours in one day, so this term here is actually just the number one because 24 hours and one day are the same thing, so I'm not changing the number 30 days.

02:30 This is the way to convert it to units by not changing the number but multiplying it by different sorts-of the number one.

02:36 So multiply by 24 hours per one day.

02:38 We can multiply by 60 minutes in one hour and then we can multiply by 60 seconds in one minute.

02:50 And the reason this works so well is that I cancel days, I cancel my units of hours, I cancel my units of minutes so that if I take 30 times 24 times 60 times 60, I will have my time in units of seconds.

03:04 So if we do this and you can do this on your own using a calculator or anything else we have, something that is approximately, 2.6 times 10 to the 6th seconds.

03:15 So that's a lot of seconds, so eventhough we're applying a small force, we do get a long time over which this force is applied so maybe we will get this spaceship to move pretty well after all.

03:23 So let's see, the second thing we have to do is find the acceleration.

03:27 Here's what we're going to need to apply Newton's second law, we know that F the force on our object which is one Newton is equal to the mass of the object times its acceleration, so we can find the acceleration quite easily.

03:39 We just take the force and divide by the mass.

03:42 In our case, our force is one Newton and our mass is 1.000 kilograms.

03:49 Remember our units of Newtons, what we have is kilograms meters per second squared divided by kilograms so this is equal to one divided by 1.000 which is 10 to the minus 3 using our scientific notation, meters per second squared which are the units for acceleration.

04:09 So now that we have these two things, our acceleration from this sort of analysis and our time from this unit analysis, we can plug in and find the distance of the change in position of our object.

04:21 Doing that we now have the change in position, X minus X0, where the initial position is zero equals 1/2 times the acceleration 10 to the minus 3, times the time which is this complicated expression 2.6 times 10 to the 6th seconds squared.

04:41 So we have to do one last thing here which is to square this quantity 2.6 times 10 to the 6th so this is 1/2 times 10 to the minus 3 and 2.6 when you square it, is approximately 6.760 but as I've already discussed, sometimes you're going to want to just make some approximations on your own, to work through a problem very quickly.

05:07 So for example in this problem, we could've taken this 2.6 and just try to put in 3 because we know very quickly 3 squared is 9.

05:14 So just by doing something very quickly and simply we can sometimes simplify our analysis and find out what the correct answer, in an exam setting especially could be without having to deal with any tricky numbers like this ones.

05:25 So the 6.760 times, and then we have 10 to the 6th squared which will be 10 to the 12th since we multiplied these exponents.

05:33 Now we do what we've always done with our unit analysis, which is to treat our coefficients as we call them, the numbers in front, on their own, so we have 6.760 and then divide it by 2 and then we treat the powers of 10 on their own, so 10 to the 12th times 10 to the minus 3, subtract 3 from 12 giving us times 10 to the 9th then all we have to do is simplify this slightly and get 6.760 divided by 2 which is about 3.380 times 10 to the 9th power, and this is in meters.

06:08 So 3 times 10 to the 9th, 10 to the 9th or changing our units, it'll be 3.380 times 10 to the 6th kilometers if we wanted to write it in kilometers.

06:21 Either one of these will be fine so be careful in an exam setting to find out which options have meters and which ones have kilometers and make sure you get your powers of 10 correct.

06:28 And so we see just from this simple example that even a tiny force if applied over a long enough period of time, these accelerations will build up more and more and more until you in fact get great distances from very, very small forces.

06:41 The equations of motion which we just used in this problem, are often referred to simply as kinematics, how things move.

06:48 On the other hand, forces and we have things under the influence of different forces.

06:53 We call dynamics because things are dynamic, they're moving.

06:57 In the problem we just solved, we saw that the link between these two and the way to solve each one was acceleration.

07:04 We want something to do with the equations of motion.

07:06 We didn't know the accelerations, so we appeal to Newton's second law F equals ma to find what the acceleration was and to plug that acceleration into the equations of motion.

07:15 In some problems, you will know the forces and be able to solve for the acceleration as we just did and put that into the equations of motion.

07:22 In other problems it will be exactly the opposite.

07:25 You will know how something move or be told how something moves and you will need to deduce the forces on those objects based on the equations of motion going to the dynamics, using again the link of acceleration.

07:36 So be aware that this is very common for problems that you are able to find the acceleration and use the acceleration to solve for the other, either the dynamics or the kinematics the equations of motion or Newton's second law, using the acceleration.

The lecture Newton‘s 2nd Law: Example, 1 Dimension by Jared Rovny is from the course Force.

### Included Quiz Questions

1. 1720m
2. 1820m
3. 2000m
4. 17.20m
5. 1530m
1. a = 243 m/s2 and d = 406 m
2. a = 406 m/s2 and d = 243m
3. a = 234 m/s2 and d = 604 m
4. a = 2.43 m/s2 and d = 4.06 m
5. a = 2430 m/s2 and d = 4060 m
1. a = 9.86m/s2 and d = 105.87 m
2. a = 98 km/s2and d = 105.87 km
3. a = 98.6m/s2and d = 10.587 m
4. a = 986m/s2and d = 762 m
5. a = 9.86m/h and d = 76.29 m

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