I've been searching for answers online about the energy involved in pushing an immoveable wall i.e. object of infinite mass, but have not found any of them to my satisfaction as of yet. I stumbled upon this forum and it answered another of my questions quite well (about the energy involved in destructive interference), so I thought I'd try here. Okay, so most people online ask a question about pushing against a wall of infinite mass and asking where the energy goes. Since the wall doesn't move, the energy involved in pushing the wall does no work and hence no energy is consumed there. So others ask what happens to the chemical energy involved since one would grow tired pushing that wall, and most answers revolve around it being converted to mechanical energy to move the muscles and heat energy dissipated, which corresponds to why one would feel hot or tired. This is all very acceptable to me, but I have a further question. I want to know if pushing an immoveable wall actually generates more heat energy loss than if the same energy was used to push one of finite mass. I thought that I should change the 'pusher' in this case to another more predictable source like a vehicle, say a tank or a car maybe. Using 2 cases would provide a clearer view of my question. In case A, let's have my tank of 1kg mass be placed right next to and touching another 1kg mass (the wall) - I will be using arbitrary values everywhere because I usually like to keep things as similar to the real world as possible - and the electric motor in the tank has a constant power of 11W. It does however have a constant 1W heat loss, giving it an effective power of 10W. By simple physics we can assume the 2 systems to be 1 mass of 2kg, and with the motor started the 2kg mass should accelerate until the friction acting on the 2kg mass increases to some arbitrary value (which I'm not really sure how to calculate) where all the energy from the motor is used to counter friction. The 2kg mass then moves at constant velocity from then on. I believe the energy involved looks something like this: 10W = 10 J s^-1 Every second 10J is supplied to the 2kg mass which is used to give it kinetic energy and overcome any friction involved: 10 J = 0.5mv^2 + some integral of friction and distance travelled during the 1 second For case B, I would have the same 1 kg tank with the same 11 W motor with the same 1 W heat loss placed right next to and touching the wall of infinite mass. The motor is started and of course, nothing moves. Here I assume that the static friction between the caterpillar tracks of the tank and the ground is high enough such that they aren't moving either, so no extra heat energy is lost there since nothing is rubbing against each other. No energy is used to overcome friction or give the system kinetic energy, so where does the 10W of power go? Does all of it get dissipated as extra heat loss i.e. full 11W of heat loss? Where would that heat loss come from? Directly from the engine? But no additional parts are moving, the wires/cables in the engine haven't had their resistances increased, so I'm very lost here. Also, I assumed a vacuum for this so please ignore sound energy or something along those lines. Am I missing something here? Please let me know. Or does a engine really get - to put it loosely - hotter when it's actually doing no work? Thanks!
Power=Work/Time. Engine output is generally stated in the U.S. in terms of Horsepower (HP). 1 HP=550 ft-lb of work per second. Work=Force*Distance HP=Torque*RPM/5252 So let's say your engine produces 10 HP@1,000 RPM. That means that at 1,000 RPM your engine produces 52.52 lb-ft of torque. In other words, if your engine was at 1,000 RPM and it was running at wide open throttle and not increasing or decreasing RPM, the load would be 52.52 lb-ft. If the load is decreased the engine would increase RPM, and if the load was increased the engine would decrease RPM. So let's say your tank against the wall produced 10 HP at 1,000 RPM (just easy numbers for this exercise). In order to be producing 10 HP at 1,000 RPM the engine must be running at 1,000 RPM and have a 52.52 lb-ft load. Well we know the engine is running and the crankshaft of the engine is rotating at 1,000 RPM, and we know that the TRACKS are not rotating, so we know the drive line is not a direct coupling. There is motion being converted to friction producing heat. That friction is causing the load on the crank, while at the same time is decoupling the crank from the tracks. If the drive line was a direct coupling, the crankshaft would rotate and the tracks would rotate at a rate directly proportional to the total gear ratio of the drive line. If the drive line is not a direct coupling, such as your automatic transmission car, the coupling between the crankshaft and transmission is accomplished via a fluid coupler, or torque converter. A torque converter makes it possible for you to sit at a traffic light with your brakes applied while your engine remains running and transmission remains in gear, and not in neutral. Surely if you try to sit at a light in an standard transmission car while remaining in gear and the clutch coupled the engine would stall, because that would be a direct coupling of the engine crank to the wheels, and if the wheels are not moving neither can the crank rotate. So in your scenario the tank is against the wall, engine running, and tracks not rotating, and the engine must have a load on it of 52.52 lb-ft in order to be producing 10 HP at 1,000 RPM. The load is coming from friction created in the fluid coupler, one side of the coupler not in motion (output of coupler), and the other side of the coupler connected to the crank (input of coupler), rotating at 1,000 RPM. The fluid between the input and output is creating friction between the two and is heating up. Automatic transmissions that use torque converters are notorious for producing large amounts of heat due to the scenario I described. It's actually the opposite, an engine gets hotter the more work it is doing and the more HP it produces. It's why your car warms up faster in the winter if you drive it, vs letting it idle in the driveway. Like I stated earlier, the more work the engine is doing the more HP it is producing. More power means more fuel is being burned per interval of time. So, if you have an engine in your car running at say 3,000 RPM, cruising down a level road at a constant speed, the engine is doing very little work. Your throttle will only be open a small amount, and the torque on the crank is low. If your engine was placed on a dyno and it was tested to be capable of producing 500 lb-ft of torque at 3,000 RPM at wide open throttle, that would be 285 HP. But while you are cruising down the road at a part throttle at 3,000 RPM the load is low, the throttle is low, and the torque is low, which means the HP is low, and the fuel consumption is low. The more load you place on the engine at 3,000 RPM the more you have to open the throttle to maintain that 3,000 RPM, and the more torque, the more work, the more HP, and the more fuel consumption, which all equate to MORE HEAT.
Having \(m=\infty\) renders all your equations useless because it isn't taking a numerical value. \(\infty\) is not an element of the Reals and thus you cannot expect any equations to make sense if you use such a 'value'. As for people getting tired when they push against walls it is due to our muscles doing work on a cellular level in order to cause our muscles to contract in the right manner. After a while this chemical process begins to build up waste products and drain energy sources in the cell, which we feel as tiredness.
Exactly. It's just the physical structure of muscle fibers "ratcheting" themselves into position, and constantly doing so during flexion. @icyveins7: You can think of our muscles as motors in a drag line that slips under load and needs constant fuel to keep its tension. The bones, on the other hand are structured for static loads, so they would not produce a sensation of effort or strain. For this reason alone, the problem has nothing to do with physics. However you could restructure your problem by leaning a heavy inert object, like a log, or a car, up against the wall. You will quickly conclude that no work is being done once the hoisting is finished. All that happens is a static loading, in which forces equalize across all the distributed members of the wall. Some may bend (or crack) under strain, so there may be a drift over time as to the exact way force is distributed. But under ideal conditions, it all comes to rest, and that's the end of any energy expenditure. Also, you can bind the shaft of a motor, switch on the power, and it will burn energy. Depending on the type of motor it may even self destruct. But that's functionally equivalent to placing a short across a power supply. You are simply wasting energy, converting it to heat.
