Understanding Newtons vs Joules - Beginner

1. Jun 4, 2015

RagnarTeller

I'm trying to understand these basic concepts and I want to understand them deeply. In learning about newtons and joules, I have a couple of questions. Please bare with me as I seek clarity.

1 netwon is the amount of force it takes to accelerate 1 KG at the rate of 1m/s^2.

If a 1KG mass of iron is floating at rest in space, and I push on it with my finger for one second to achieve a velocity of 1 meter per second, I've applied 1 newton of force. That's easy enough.

Confirmation #1: What if I push 2x as hard for 1/2 of a second or 100x as hard for 1/100 of a second. In other words, the acceleration can happen over an entire second or all at once, as long as the cube is going 1m/s when measured at 1 second after the force begins being applied. This is all still 1 newton, correct?

Confirmation #2: If a 1KG mass of iron is traveling through space at 100m/s, and I apply 1 newton of force with my finger in the direction of travel, will that mass then be traveling at 101m/s?

Confirmation #3: If I apply 1N to the opposite direction of travel, that mass will then be traveling 99m/s, correct?

If a mass is floating at rest in space, and I apply 1N over 1meter (1joule):
Question#1: How fast will that mass be traveling at the 1meter mark?
Question#2: How long will it take to reach the 1meter mark?
Question#3: Why measure joules in distance instead of time, like the newton? It seems like the same thing to me. If you apply 1N for 1m, you will still accelerate that mass to a final velocity over a period of time, which seems like you would measure in newtons applied to the mass. I just don't get why we need the joule.

2. Jun 4, 2015

Staff: Mentor

What was its velocity at time zero?
What was its velocity after 1 second?
What was its average velocity over the 1 second?
How far did it travel at this average velocity over the 1 second?

Chet

3. Jun 4, 2015

RagnarTeller

At time zero, v=0
Velocity after 1 second would be 1 m/s
The average velocity would be .5 m/s ?
It would have only traveled .5 meters?

4. Jun 4, 2015

Yes and yes.

5. Jun 4, 2015

EM_Guy

No. The force that you exert is not dependent on the time you spending exerting that force. If you push something 2x as hard, then you have doubled your force. If you push something 100x as hard, then you have increased your force by 100x. 1 N = 1kg * 1 m/s^2.

It depends on how long you are applying that force. F = ma. So, if a 1 kg mass (doesn't matter if it is iron) is traveling at 100 m/s, and you apply 1 N of force in the direction of travel, then the mass will accelerate for as long as you are applying that force to the object. The acceleration would be a = F/m. In this case, a = 1 N / 1kg = 1 m/s^2. Now, once you find the acceleration due to your force, you can forget about your force. Now you are doing kinematics. If an object is moving 100 m/s and it is accelerating 1 m/s^2 in the same direction of its travel, then what will be its velocity after 1 s, 2 s, 3.48 s, 100 s, 423 s? The more time that the object is accelerating, the more its velocity will change. If the acceleration is constant, then your change in velocity is your acceleration x time.

6. Jun 5, 2015

dean barry

A handy equation for velocity change problems is derived from Newtons famous: f = m * a
( f = (constant) force in Newtons, m = mass in kgs, a = (constant) acceleration in (m/s)/s )
you also know that a = velocity change (vc) / time (t)
(velocity in m/s, time in seconds)
so, you get: f = m * ( vc / t )
transpose for vc:
vc = (f * t) / m

intersestingly, the force applied * time applied = the change in momentum

7. Jun 5, 2015

Staff: Mentor

That's its definition, but not its only application. You can also apply it to a wall by pushing on it and to a floor by standing on it. Neither involve motion.

8. Jun 10, 2015

RagnarTeller

Wow. Thanks for your help everybody! This is definitely the best discussion forum on the internet.

Would it be 101 m/s , 102 m/s, 103.48 m/s, 200 m/s, and 523 m/s? Just double checking.

That's a great point russ. I get it intuitively, force can exist without acceleration, but how do we reconcile that observation with f=ma? It seems like there's something missing. If vc = (f * t) / m, and you push on a wall with 10 newtons of force for 10 seconds, and it's acceleration is 0, what is the mass? Would it be infinite?
10N*10sec/m = 0vc

Is there a commonly used analogy if the physics world for rookies like me? Maybe: Money=Force and Dollars=Newtons. It doesn't matter if you pay for 1oz of gold all at once or get on a payment plan. The dollars still have the ability to do a certain amount of work, in this case, $1200 can purchase 1oz of gold. The$1200 would be force, the gold would be mass and "purchase" would be acceleration as it moves ownership from one place to another.

If that analogy works for you guys, could you think of how joules would fit into that analogy, or would it just not work?

9. Jun 10, 2015

Staff: Mentor

The way to understand Newton's 2nd law is: ΣF = ma. If something doesn't accelerate, it just means that the net force on it is zero.

Not quite. Realize when you push on a wall that yours is not the only force acting on the wall. (It is presumably attached to some structure.) The net force on the wall will remain zero.

10. Jun 10, 2015

SteamKing

Staff Emeritus
If the wall is not moving as the result of pushing on it, you can't say anything about its mass.

F = ma, or rather Fnet = ma applies to objects in motion. If the wall cannot move, then by definition a = v = 0, and the wall can be any mass. Since the wall is not moving, then that implies that Fnet = 0, which means that if someone pushes on the wall with force F, then the wall pushes back with an equal and opposite force -F, so that Fnet = F - F = 0.

This is stated in Newton's Third Law of Motion:

http://en.wikipedia.org/wiki/Newton's_laws_of_motion

11. Jun 10, 2015

A.T.

It applies to all objects. It just isn't useful for some purposes in some cases.

The equal and opposite force -F doesn't act on the wall, so it has nothing to do with Fnet on the wall.

Last edited: Jun 10, 2015
12. Jun 10, 2015

EM_Guy

In this case, yes, because the acceleration is 1 m/s^2. In general though, acceleration is not equal to 1 m/s^2.

For example, a car that goes from 0 mph to 60 mph in 10 seconds experiences an average acceleration of 60 mph / 10 s or 6 mphps. On average, every second, he is increasing his velocity by 6 mph.

Working with this example, it would be a good practice to do some graphing. If someone is accelerating 6 mphps for 10 s, what is his velocity as a function of time (if he starts from an initial velocity of 0 mph)? What is his displacement as a function of time? What does the area under the acceleration versus time graph represent? What does the slope of the velocity versus time graph represent? What does the area under the velocity versus time graph represent? On the displacement versus time graph, what does the slope of the tangent line (for any given time) represent? What does the acceleration versus time function look like? What kind of function is the velocity versus time graph? What kind of function is the displacement versus time graph? If you can graph all of these functions and answer all of these equations, you will have a solid understanding of 1-dimensional kinematics. (And then we can start talking about 2 dimensions - like the trajectory of a bullet in a gravitational field.)

13. Jun 10, 2015

RagnarTeller

Ok, I think I'm understanding Force and the Newton. Now to understand the difference between Force and Energy. Let me digest this discussion, from a previous thread, for a bit.

14. Jun 11, 2015

RagnarTeller

Ok, I think I have it.

The difference between force and energy is that a force is what does work; energy is the work that is done. It's kind of like the difference between money and spending money.

Dollars don't do anything, you have to spend them. The act of using dollars to transfer goods is like the act of using force to accelerate mass.

Newtons are like dollars, Joules are like the price you pay to purchase something.

Please tell me if you think this is an accurate description of the difference between force (newtons) and energy (joules).

15. Jun 11, 2015

EM_Guy

Force is not a conserved quantity. Energy is. Money would be more like energy (if the government stopped printing it and if inflation never happened, but I digress). The point is that you don't "use up" force. You use up energy.

As an example, when you lift a barbell from the ground to overhead, you do work on the barbell. If we consider the ground as a reference point, then the potential energy of the barbell on the ground is 0 ft-lb. As you lift the barbell (performing work on it), you increase its potential energy (because it is getting higher). The total work that you perform on the barbell is equal to the increase of potential energy of the barbell. So, if you lift 100 pounds overhead, and if overhead for you is 6 feet, then you have done 600 ft-lb of work, and the potential energy of the barbell at that point is 600 ft-lb.