# Two Basic Physics Questions

1. May 29, 2005

### Hookflash

I'm an adult doing the self-education thing, having dropped out of high-school (big mistake, incidentally). I have a couple basic physics questions that I can't seem to find the answers to:

1) If F=ma, why do all objects the same distance from the earth's surface accelerate at 9.8 m/s^2? Shouldn't objects of greater mass accelerate more slowly, according to Newton's second law (assuming Earth's gravity is constant)?

2) W=Fd (where W="work"), but, when I lift an object 4 meters at a constant speed I am said to have done twice as much work as if I had lifted it 2 meters at a constant speed. This seems strange to me, since, in both cases, the initial acceleration of the object (as it goes from being at rest to moving at a constant speed) is the only non-zero force involved. Since the distance after that is travelled at a constant speed (F=m(0)=0), shouldn't it add nothing to the total force required to lift the object?

I know, intuitively, that there is something wrong with my reasoning in both of these questions, but I can't quite figure what it is. Help would be appreciated.

2. May 29, 2005

1) Its a bit of a circular argument but this might help; a=F/m for all bodies, and the F term for a falling body is its weight. Weight = m*g so F is proportional to m for all bodies.
In other words, bigger bodies have bigger forces on them, and this bigger force has to act on a bigger mass... so thay all accelerate at the same rate!

2) You are doing work continually whilst lifting a body - it is gaining Gravitational Potential energy. You are mixing up Newton's first law here. You are continually applying a force as you lift an object, to counteract gravity. So, you are doing work.

If the body was moving on a frictionless surface parallel to the Earth's surface at constant velocity, yes the Force would be zero and no work would be done.

Oh, and by the way, Welcome to Physics forums..

3. May 29, 2005

### whozum

1) The universal gravitational force is given as

$$F = \frac{GM_1m_2}{r^2}$$ , where G is the gravitational constant (~6.67e-11)

For the earth we can assign values to the above equation and get it to a more understandable equation:

$$F_{grav} = \frac{GM_{earth}m}{r_{earth}^2}$$

As you quoted newtons law is F = ma, solving this for acceleration we get a = F/m, and so

$$a = \frac{F}{m} = \frac{\frac{GM_{earth}m}{r^2}}{m} = \frac{GM_{earth}}{r^2}$$

This is the final expression regarding acceleration due to gravity. Notice in this expression, there is no mention of the objects mass. The acceleration of an object due to the gravitational field of the earth does not depend on its mass.

You can also find this by setting the two forces equal. The gravitational force is as given above, and Newton's second Law says that the net force of an object is given by its net acceleration multiplied by its mass. Since gravity is the only force acting on the body, then the net force is just the gravitational force:

$$F = ma = \frac{GMm}{r^2}$$

Notice m occurs on both sides, and it cancels giving our original result

$$\frac{F}{m} = a = \frac{GM}{r^2}$$

Finally calculating the actual values we can show that g = 9.8

$$a = \frac{GM}{r^2}$$

G = 6.67 x 10^-11
M = 5.98 x 10^24 kg
r = 6.378 x 10^6 m

$$a = \frac{(6.67\times 10^{-11})(5.98\times 10^24)}{(6.378\times 10^6)^2} = 9.805234578 \frac{m}{s^2}$$!

4. May 29, 2005

### Hookflash

Adrian Baker: Thanks for your reply. If I understand you correctly, Earth's gravitational pull varies according to the mass of the object it's acting on?

With question #2, I confess I'm still a tad bit confused. Doesn't a non-zero force always cause acceleration? For example, say I apply an upward force of 10 Newtons on a 1 Kg object, while Earth is applying a downward force of 9.8 Newtons on that same object. Shouldn't that mean there's now a net upward force of 0.2 Newtons acting on the object? And, if so: F=ma, 0.2 = 1a, a = 0.2 m/s^2? This is the confusing point for me. Why doesn't the object accelerate upwards when I apply a constant force to lift it?

5. May 29, 2005

### Hookflash

Whozum: Ahh! Things are starting to click (sort of). Thanks. :)

6. May 29, 2005

### whozum

The error you are making here is the following line:

.

Remember that you are lifting the object. What are you lifting it against? Gravity ofcourse! You are doing work AGAINST the gravitational force, so for you to move it against hte gravitational force, you must provide a force greater than that of gravity. The total work against a gravitational field is given as the strength of that field multiplied by the object's mass and its component of displacement parallel to the field.

Or algebraically as you said,

$$W = F \times d$$

F is just mg and d is the total displacement $(x-x_0)[/tex], so $$W = mg(x-x_0)$$ Now we can play with it a bit to show an important point, if you distribute mg in the above equation you get $$W = mgx - mgx_0$$ with simple algebra. What do you notice about it? It loosk alot like the (simplified) equation for potential energy doesnt it? PE = mgh. Our equation for W is just equal to the difference in potential energy for the two points, and that makes sense. When you do work against the gravitational force, you are simply changing the state of potential energy of the object. The amount of work you have done is the amount of potential energy you have displaced! This is why work and energy are so analogous in cases like this (and many others). edit: A point from above, only the component parallel to the field is what matters, and this is shown mathematically by the dot product, work is defined as $$W = F \cdot d$$, which equates to $$W = Fdcos(\theta)$$ where [itex] \theta$ is the angle between the gravitational field and the path of movement. If you are moving completely parallel to the field (only up and down) then the angle between the field and the path of travel is 0, and [itex] cos(\theta=0) = 1 [/tex] and W just becomes Fd. If you are moving completely perpendicular to the field (side to side) then the angle between the field and the path is 90 degrees, and [itex] cos(\theta=90) = 0 [/tex] and the work equation becomes W = Fd*0 = 0. This is why no work is done against gravity when you rae moving from side to side. If you compare this to my explanation between work and energy states above, you will see why.

Last edited: May 29, 2005
7. May 29, 2005

### whozum

The force varies based on the mass, but the acceleration is always the same.

A nonzero NET force always causes a NET acceleration. This is exactly what Newton's second law says. If you did apply a constant 10N upward force, then the object WOULD and does accelerate upward at 0.2m/s^2. You may not recognize it as acceleration because you are a) not providing a constant force and b) can only lift it so long before your at an arms length.

8. May 29, 2005

### Hookflash

whozum & Adrian: Thanks for the help, guys. It's finally making sense. Now I can continue with my frustratingly terse book (Basic Physics: A Self-Teaching Guide). :)

9. May 29, 2005

### whozum

No problem! I hope you will return with any more questions. (I really do, I'm bored to death).

10. May 30, 2005

### Hookflash

Ok, I think I may have discovered what had me so confused all this time. In the physics book I'm currently reading, they give the equation for "work" as W=F*d, but then define F as the net force acting on some object (upon which work is done) over d! But, shouldn't F really be only the force exerted by the agent doing the work?
For example, when I lift a 1 Newton object at a constant speed, the net force is zero, but the force I'm exerting is 1 Newton. If I'm right, then I have to say: This was one of the most confusing mistakes I've ever encountered in a physics book.

11. May 30, 2005

### Order

Mistake by you...or I believe you mean the book. I have to agree the concept of work is really contra-intuitive. Maybe someone can explain to me as well?

You are right that you should only use the force exerted by the agent in your case of overcoming the gravity potential. However the book would be correct in the case where there is no field to overcome. For example if you have a tug-of-war and one team is constantly winning so that they move at a constant speed in their direction. (An unstable situation that won't very often happen in reality.) Then the both teams apply the same force, so there is no net force. Then they don't do any work, although they will be very tired. (But if they had cooperated instead of combatting each other they could have got some work done .)

So my question is if anyone has a clear definition of the force in the formula $$W = F \cdot d$$?

12. May 30, 2005

### whozum

Well I'll disect this and see where it gets me. Work (potential energy change) is defined as a line integral of a force (usually a force field) over the path of motion, or

$$\int_{r_1}^{r_2} \vec{F} \cdot d\vec{r}$$

Over an infinitesimally small change in path, an infitesimally small force is applied such that when taken in perspective, each infinitesimal (sp) is equal in magnitude to all others, and thei ntegral is taken to find the exact value over the region.

In the case of something like gravity, then given the force equation we can integrate to find the change in potential energy:

$$F_{grav} = \frac{GMm}{r^2}$$

$$W = GMm \int_{r_1}^{r_2} \frac{1}{\vec{r}^2} \cdot d\vec{r}$$

Where W as we mentioned before represents the work, and change in potential energy. The evaluated integral comes out to

$$W = -GMm \left(\frac{1}{r_2} - \frac{1}{r_1}\right)$$

Thus we can interpret the above as the work done by (or on) a gravitational field is given by the product of GMm multiplied by the difference of the inverses of the distance. Note the negative sign and its indication that the default interaction is an attractive one. There is work done since there is a change in potential energy (it has been transferred to kinetic energy).

In your case though, you are saying that you are slowly lowering a book at a constant velocity, such that the net force on the book is zero throughout. This is really a tough call and I'm not sure I can give the best answer, but my proposition is that the work done by the gravitational field AND the work done by you (yes, you can do work) sums up to zero. With respect to the gravitational field (and ignoring your presence), the total work done by the gravitational field can be given by the above equations. Perhaps it is in this aspect that the points you give and the book's points differ.

13. May 30, 2005

### Staff: Mentor

Sloppy writing in a physics book makes me nuts. So I feel your pain.

You are correct. The work done by a force F acting on an object will equal F*d, where d is the displacement of the object in the direction of the force.

You are right, of course. If you lift the object at constant speed for a distance of 1 meter, then you have done F*d = 1 N-m = 1 Joule of work.

Of course gravity is also exerting a force of 1 N on the object (acting down, so negative). So the work done by gravity is F*d = (-1 N)* (1 m) = - 1 Joule. (Work is negative if the object moves opposite to the direction of the force.) So you can say that the total work done by all forces is zero. Which makes sense, since the total (net) force on the object is zero.

14. May 30, 2005

### Order

Yes, it seems that whozum and Doc Al are right. And I was "wrong". The total work done is zero. Because according to the definition of work it is the change in kinetic energy. (Including rotational energy. This is the so called work-energy theorem.) And Hooklash moved the object with a constant speed so that it didn't change => The kinetic energy didn't change => The work done is zero.

This also explains what is wrong about how the subject is treated in school books. Many books use a correct language, but few discuss the difference between total work done and the work done by someone/something. I can understand teachers however. Teaching physics is not easy.

/Order

Last edited: May 30, 2005
15. Jun 4, 2005

### paulom2001

That is Order. When we´re talking about WORK generally we want that one done by an specifically agent instead of the NET WORK. In this case we just need that specifically force, instead of the net force.

16. Jun 4, 2005