# Pushing wall

• B
Dear PF Forum,
I have a problem with calculating energy which I should have learnt long ago in my high school time.
If I push (accelerate) a 1 kg object 1 meter/sec2 for 10 seconds, I spend energy like this.
Distance = 1/2 at2. So I would have pushed this object for 50 metres. So I'm using energy 50 metres * 1 kg * 1 m/sec2 = 50 joules.
Am I right?
Now for the second question. Supposed I push a wall with the same force as above for 10 seconds, how can I calculate the energy consumption? Considering the wall doesn't move.
Thank you very much.

mfb
Mentor
Am I right?
Yes.
Supposed I push a wall with the same force as above for 10 seconds, how can I calculate the energy consumption? Considering the wall doesn't move.
Repeat the same calculation with a=0.

Your body will actually need some energy because it is not 100% efficient (and it needs energy just for standing around as well), but that is a biology question.

Yes. Repeat the same calculation with a=0.
Okayyy, so.
J = n.m
J = Kg * a * m
J = 1 * 0 * 0
I don't spend energy?
I'm sorry I can't understand it. It's like imagining 4 dimension space time in 3 dimension space.
Don't I spend any energy at all?
Perhaps yes/no will be sufficient, while I'm thinking my next question, based on that scenario.

Thank you very much mfb.

mfb
Mentor
I don't spend energy?
Let's say you do not give energy to the wall.
Your muscles actually spend energy, but only to support your body in its position, you convert all the energy to heat and nothing to mechanical energy.

• praveena and Stephanus
The way our muscles work is that only some of the fibers are contracting at any given moment. This helps forestall global exhaustion of the muscle. It is the contracting and sliding past one another done by the fibers that is the mechanical work for this system. So even though the global muscle does not move when pushing against a stationary object, the fibers themselves are doing a lot of moving. This gets transformed into heat and chemical waste products.

So I guess you could calculate the total energy expended if you knew things like the length of contraction and expansion for the individual fibers, the friction between fibers, the efficiency of the contractions, etc. You will not get there, as you found out, by looking at the global motion of the muscle.

• Stephanus and mfb
You do expend energy, but not it a way that can be calculated by physics equations. As mentioned above, it's a biology question. The energy expended would show up as added body heat and food resource consumption.

• Stephanus
A.T.
I spend energy like this
You are computing the work that you did on the object, not the energy you spend.

• Dale
Thanks @mfb
Thanks @mfig
Thanks @OldYat47
Thanks @A.T.
There's energy but not in the form of physics but biology. I'll reread Newton again.

Have you ever tried holding a car steady on a hill by slipping the clutch? The engine was burning a lot of fuel, but the wheels were doing no work against gravity.

• mfb
energy spent=work done against friction of floor= large friction * small displacement of feet

• Stephanus
Have you ever tried holding a car steady on a hill by slipping the clutch? The engine was burning a lot of fuel, but the wheels were doing no work against gravity.
I'll think about it. I would HAVE liked to ask about how much energy spend for holding the car steady, but I'll study Newton first.
Thanks.

A.T.
There's energy but not in the form of physics but biology.
The energy is dissipated as heat.

• Stephanus
The energy is dissipated as heat.
Thank you very much. This answer helps me much.
Actually I want to know why in rocket if we keep pushing 1 newton per meter second, the watt increases. But that for another thread. I do want to delve into this deeper.

Thank you

A.T.
Actually I want to know why in rocket if we keep pushing 1 newton per meter second, the watt increases.
You can strike all that "per.." there. The power is frame dependent. It increases in frames where the velocity increases.

• Stephanus
Dear PF Forum,
I have a problem with calculating energy which I should have learnt long ago in my high school time.
If I push (accelerate) a 1 kg object 1 meter/sec2 for 10 seconds, I spend energy like this.
Distance = 1/2 at2. So I would have pushed this object for 50 metres. So I'm using energy 50 metres * 1 kg * 1 m/sec2 = 50 joules.
Am I right?
Now for the second question. Supposed I push a wall with the same force as above for 10 seconds, how can I calculate the energy consumption? Considering the wall doesn't move.
Thank you very much.

Energy spent in pushing the wall releases in other forms like, heat energy, calories burnt in the body which comes in the form of sweat from the body

• Stephanus
mfb
Mentor
You can strike all that "per.." there. The power is frame dependent. It increases in frames where the velocity increases.
And it decreases in others.
Energy conservation works for rockets, but you absolutely have to take into account the kinetic energy of the exhaust, which makes the analysis messy. Conservation of momentum is easier to use.

• Stephanus and Dale
You can strike all that "per.." there. The power is frame dependent. It increases in frames where the velocity increases.
Thanks @A.T. , I don't understand this, but I'll find it out myself so that I can understand it. I'll try to work it out.

Energy spent in pushing the wall releases in other forms like, heat energy, calories burnt in the body which comes in the form of sweat from the body
Yes, someone above... oh @OldYat47
You do expend energy, but not it a way that can be calculated by physics equations. As mentioned above, it's a biology question. The energy expended would show up as added body heat and food resource consumption.
and @mfig said that.
But how much energy spent if if push a wall for 10 seconds with the same force as we move a 1 kg object for 10 seconds? But, just let it go, actually I want to know, why watt increases with time in Newton. I'll work it out myself.
THanks @Gudiya kumari
And it decreases in others.
Energy conservation works for rockets, but you absolutely have to take into account the kinetic energy of the exhaust, which makes the analysis messy. Conservation of momentum is easier to use.
Yes calculating kinetic energy is simple.
If a 1000 ton rocket fires 1 miligram ejecta behind for 100 km/s each second (I would like to use 100 thousands km/second, but it will be relativistic, and it will add another problem for me) so the rocket will be accelerated... 0.1 micrometer per second per second, (Newton third law, right?) and for 1 milion seconds it speed will be... 0.1 meter per second so the kinetic energy would be around... 5000 joules? But I've been thinking this thought experiment before I do post this in PF Forum. Actually, I want to understand this, but I haven't been ready to ask this question until I really understand this Newton thing.
Thank you very much ladies and gentlemen.

mfb
Mentor
Yes calculating kinetic energy is simple.
If a 1000 ton rocket fires 1 miligram ejecta behind for 100 km/s each second (I would like to use 100 thousands km/second, but it will be relativistic, and it will add another problem for me) so the rocket will be accelerated... 0.1 micrometer per second per second, (Newton third law, right?) and for 1 milion seconds it speed will be... 0.1 meter per second so the kinetic energy would be around... 5000 joules? But I've been thinking this thought experiment before I do post this in PF Forum. Actually, I want to understand this, but I haven't been ready to ask this question until I really understand this Newton thing.
Thank you very much ladies and gentlemen.
The kinetic energy of the rocket is 5000 J, but the kinetic energy of the projectiles is 5 GJ. You can't look at the kinetic energy of the rocket without ignoring the massive amount of energy that goes into the projectiles. The first projectile alone gets about 5 kJ. The last one gets about 4.9999999 kJ. A bit less - while the rocket gains a bit more kinetic energy there.

• Stephanus
The kinetic energy of the rocket is 5000 J, but the kinetic energy of the projectiles is 5 GJ. You can't look at the kinetic energy of the rocket without ignoring the massive amount of energy that goes into the projectiles. The first projectile alone gets about 5 kJ. The last one gets about 4.9999999 kJ. A bit less - while the rocket gains a bit more kinetic energy there.
Thanks @mfb, that I realize, but.. I can tamper the number so that the energy kinetic of one the projectile could be negative.
I mean, from a rest observer where the rocket starts, if the rocket is fast enough, and wrt to a rest observer, one of the projectile could have the same direction as the rocket. Because in rocket's frame, the velocity of the projectile is minus, but wrt a rest observer the velocity of the projectile could be positive.
I think, ignoring heat and miniscule interstellar medium friction the energy kinetic of both the rokect and the projectile should be zero. I mean there were two walls, one wall hit by the rocket (and some projectiles if they are in the same direction as the rocket) and the other wall hit by the projectiles (the ones with the reverse direction as the rocket) so I think the energy kinetic will be the some for both wall? Except for a projectile with zero velocity.
I'm sorry, I just can't arrange my question.
Actually I'd like to study two scenarios.
A rest observer sees a rocket travels 100 km/s at the east.
And that rocket fires a half of its mass backward say 10 km/s, so the velocity of the object is 90 km/s while the velocity of the rocket is 110 km/s
From a rest observer wrt rocket, it will see the rocket (now) travels 10 km/s to the east and the projectile 10 km/s to the west.

And I still don't understand the relation of newton and watt.
Say, a stationary rocket fires a projectile backward, after the effect of the acceleration diminish (now the rocket travels constantly), say 1 second after that the rocket fires another projectile.
After the effect of the acceleration diminish (now the rocket travels constantly), say 1 second after that the rocket fires another projectile.
After the effect of the acceleration diminish (now the rocket travels constantly), say 1 second after that the rocket fires another projectile.
and on and on,... why Watt increases over time?
I know, Newton doesn't work that way, to calculate Energy we use J = N * m. In above scenario, the acceleration only felt for each second.

Say, to fire a projectile backward in V velocity, the rocket needs W watt (is this a correct statement)?
Now, after 1 second, the effect of the acceleration diminish, the rocket fire another projectile with V velocity that neewds W watt, of course along the journey the power doesn't increase, right?
Or I could have edit my previous statement.
Now, after 1 milisecond, the effect of the acceleration diminishes, the rocket fire another projectile with V velocity that neewds W watt, of course ...
Now, after 1 microsecond, ..., the rocket fire another projectile with V velocity that neewds W watt...
I could have wrote 1 pico second, and from up here it looks like dt, constant acceleration. Just how hard it is to push a projectile backward in T0 and in T100 seconds?
I've studied read special relativity in PF Forum for 1 year, and I realize that there's no absolute frame of reference in this universe, so what's so special about time zero, when the rocket starts travel and time 100 after 100 seconds that the rocket needs more power to push another km/s as at the first place?

By "special relativity" I mean constant speed in 1 spatial dimension.

Thanks @mfbI think, ignoring heat and miniscule interstellar medium friction the energy kinetic of both the rokect and the projectile should be zero. I mean there were two walls, one wall hit by the rocket (and some projectiles if they are in the same direction as the rocket) and the other wall hit by the projectiles (the ones with the reverse direction as the rocket) so I think the energy kinetic will be the some for both wall?
No, I think this is wrong!. The momentums are the same, but the kinetic energies are not.
And looking at their formula
##\text{Momentum } = m.v##
##\text{Kinetic energy} = \frac{1}{2}m.v^2##
It looks like Ke is the integral of Momentum: ##\int{m.v dv}##
Or, it's just a coincidence.

mfb
Mentor
Kinetic energies cannot be negative. And in reference frames where the last pellet has a higher kinetic energy than the last one, the rocket loses energy (you also have to consider its decreasing mass).
I think, ignoring heat and miniscule interstellar medium friction the energy kinetic of both the rokect and the projectile should be zero. I mean there were two walls, one wall hit by the rocket (and some projectiles if they are in the same direction as the rocket) and the other wall hit by the projectiles (the ones with the reverse direction as the rocket) so I think the energy kinetic will be the some for both wall?
No. Why do you expect that? Kinetic energy is not momentum, and it does not have a direction or separate conservation law.

If you sum the kinetic energy of projectile and rocket before and after separation, shooting it away will always increase the sum by the same amount, no matter in which reference frame you do the calculation (assuming nonrelativistic motion, otherwise we have to take the energy source of the rocket into account).

• Stephanus
Kinetic energies cannot be negative. And in reference frames where the last pellet has a higher kinetic energy than the last one, the rocket loses energy (you also have to consider its decreasing mass).
"Decreasing mass", yes, I have considered that.
No. Why do you expect that? Kinetic energy is not momentum, and it does not have a direction or separate conservation law.
I don't know, It's just a guess. But it's wrong as you say.
If you sum the kinetic energy of projectile and rocket before and after separation, shooting it away will always increase the sum by the same amount, no matter in which reference frame you do the calculation (assuming nonrelativistic motion, otherwise we have to take the energy source of the rocket into account).
Kinetic energies cannot be negative. And in reference frames where the last pellet has a higher kinetic energy than the last one, the rocket loses energy (you also have to consider its decreasing mass).No. Why do you expect that? Kinetic energy is not momentum, and it does not have a direction or separate conservation law.

If you sum the kinetic energy of projectile and rocket before and after separation, shooting it away will always increase the sum by the same amount, no matter in which reference frame you do the calculation (assuming nonrelativistic motion, otherwise we have to take the energy source of the rocket into account).