How much energy does the force of gravity apply to a bullet?

[Hilmy atha]

Homework Statement

A bullet was shot at 40m/s
Elevation angle : 37°
Bullet mass = 0.01 Kg

Question : how much work is done by the force of gravity to the bullet since it was shot until it reaches the ground again

Homework Equations

mgh = E.Pot
1/2mv ^2 = E kinetic

The Attempt at a Solution

My teacher said, Work = F.S =0 joule, because gravity is only pulling downwards in this case, but S (distance) is in X axis, which means if we use horizontal Force, it will be zero, thus the work is Zero. If we use the vertical distance, it is zero. Thus the work is zero in both case (Work in X and Y axis)

Okay, i understand the logic in there. It suits the math

But my REAL question,
Certain amount energy was needed to shoot the bullet, correct? But somehow, the gravity make the suppossedly moving bullets back to stationary condition. Doesn't that require energy? To stop something moving? So how can the answer be 0 joule :/I have said this to my teacher but she doesn't seem to give a satisfying logical answer. Can you help me? .,.

 

[kuruman]

But somehow, the gravity make the suppossedly moving bullets back to stationary condition.

If by "stationary" you mean stopping the bullet, yes, energy is required to do that. That energy is provided by the ground, not gravity. From the time of launch until just before the bullet hits the ground, the total work done by gravity is zero. Gravity does negative work while the bullet is rising and an equal amount of positive work while the bullet is dropping. Let us know if this makes sense to you.

 

[BvU]

Good you don't take something for granted until you understand it.

Perhaps I can give you 'another' way to look at this:
Bullet goes in a projectile trajectory. In the absence of air resistance a parabolic trajectory. Let's look at that.
In the upgoing part gravity pulls down and the bullet goes up, so gravity is doing negative work: the bullet gains some potential energy.
In the downgoing part of the trajectory, the bullet loses that potential energy again, so gravity does positive work. Sum of up and down: zero.

Now about stopping a bullet, a most unhealthy activity.
As you say, when a bullet comes to a halt, its kinetic energy has to go somewhere. Ususally it is converted to heat by friction forces. A pile of sand is a good way to do that. Note that in you scenario the kinetic energy of the bullet at shooting time is equal to the kinetic energy at arrival. Exactly. (In other words: the 0 J of work from gravity is rather evident there !).

And note that gravity only affects the vertical component of the bullet velocity. But a scenario where the bullet is fired straight up proceeds the same way: without air resistance it comes down on you at the same speed it was launched. So no net work was done on it by gravity !

Finally, you can observe that things change when a projectile is launched from high up: if the target is much lower than the 'stopping place' some additional energy from work done by gravity makes its impact bigger than a projectile fired the other way, up. In the first case that work by gravity is equal the work done (potential energy provided) by the poor chap who dragged the projectile all the way up from the ground to the top of the castle tower.

 

[Hilmy atha]

That energy is provided by the ground, not gravity. From the time of launch until just before the bullet hits the ground, the total work done by gravity is zero. Let us know if this makes sense to you.

I see, that makes sense. Can you explain more about "energy is provided by ground, not gravity"
where does that energy come from anyway?

 

[kuruman]

... where does that energy come from anyway?

It depends on what you hit. If it's a pile of sand as BvU recommended, the bullet dissipates its energy into kinetic energy of gazillions of grains of sand that eventually come to rest through friction, so that ultimately the bullet's energy goes into heat. If you hit a metal sheet, both the sheet and the metal deform permanently, the bullet may partially melt, some of the energy goes into sound waves, there is also heat generated and so on.

On edit: More correctly, you should ask "where does the energy of the bullet go to?"

 

[Hilmy atha]

Good you don't take something for granted until you understand it.

Finally, you can observe that things change when a projectile is launched from high up: if the target is much lower than the 'stopping place' some additional energy from work done by gravity makes its impact bigger than a projectile fired the other way, up. In the first case that work by gravity is equal the work done (potential energy provided) by the poor chap who dragged the projectile all the way up from the ground to the top of the castle tower.

I see.. so is my drawing below correct?

 

[BvU]

I see.. so is my drawing below correct?View attachment 115530

yes

 

[kuruman]

Yes.

 

[BvU]

ou explain more about "energy is provided

Admittedly taken out of context.

This is not going well. You shouldn't associate stopping with providing energy. Stopping is absorbing kinetic energy by converting it to heat (mostly). In other words: the kinetic energy of the projectile is converted to kinetic energy of something else. Ususally smaller, e.g. a lot of grains of sand. They spread and friction (same process on a smaller scale) converts it to extra random motion of molecules (heat).

When a bullet or projectile hits something, it can do work on the target, including setting it into motion. The target does negative work (it is being pushed in the opposite direction of the force the target exercises). The projectile exercises force in ths direction it it going, so it does positive work (and thereby loses energy)

 

[Hilmy atha]

Okay thank you for your time everyone ! :D now i understand @kuruman
@BvU