When a falling object reaches terminal velocity, what is the net force acting on it?
150308
There is no net force acting on it. When there is a net, or resultant force acting on an object it causes an acceleration. As in this case a terminal velocity has been reached, that is no further acceleration is occuring, the resultant force is zero.\n\nThis is represented in the equation known as Newton's Second Law, or in Applied Mathematics, Newton Two. The equation is:\n\nF = ma and it means Force = Mass x Acceleration where the units are:\n\nNewtons (N) = Kilograms (Kg) x Meters per Second (ms^-1).
616393
Any thing moving through the atmosphere experiences the force of gravity and a drag force due to the friction of the air as it moves through it. When the object is falling, the force of gravity is directed towards the ground and the drag force directed upwards. As the body moves faster, the drag force becomes larger and larger until it equals the force of gravity and the object then falls at a constant velocity. This velocity is called the terminal velocity. For a person jumping out of an airplane, in skydiving, it is roughly 120-150 miles per hour. The terminal velocity of a water drop is much smaller. A large raindrop (5 mm in diameter) will fall at a velocity of around 20 mph. If you get hit by a large raindrop you will definitely feel it. However its still not large enough to cause much pain. The smaller the raindrop the slower it falls and hence the less energy it has when it hits you. If you dropped a grain of sand the same size as a raindrop it wouldn't hurt either. Pebbles are much heavier than raindrops and hit your head with a lot more force. Force is directly proportional to the mass and velocity of an object.\n\nOf course hail, which is really just a bunch of frozen raindrops which have joined together, is actually less dense than water but will certainly hurt due to its large mass.
913014
F(net) = m x a\n\nF (net) = F (forward) - F(air friction)\n\nso  F(air friction)  = F (net) - F(forward)\n\nMass is given 20 kg\n \nnet acceleration given 2.0  m/s^2\n\n\nSolve for the Net Force, subtract the forward force, the difference is the force due to air friction.
478612
It depends on the object.  Tests with coins have measured velocities too low to cause signifcant damage to the average human head.\n\nTerminal velocity, in this case the maximum speed a falling object will reach, depends on the shape of the object, the accelleration, and the qualities of the atmosphere.\n\nWhen an object is dropped near earth, it accellerates towards the earth at about 32 ft/sec/sec.  It goes faster and faster until the frictional drag from the air, has a slowing effect equal to the speeding effect from gravity.  From that point on it will stay approximately the same speed.  A feather has a very low terminal velocity.  A crowbar has a much higher one.  Its all about friction in air.\n\nA penny dropped near a planet without an atmosphere would accellerate toward the planet, getting faster and faster until it hit.  This scenario could be deadly for the space-suited person in the way.\n\nWater falling has an additional complication in that the air drag changes the shape of the water droplet and that in turn affect the drag.  A water droplet's changing shape on impact also affects the damage it might do a person.
811229
a=Fnet/m\n.5x1.5=0.75N net force\n\nFnet=Fa+Ff\nFf=.75-1.2\nFf=-.45\n\nI think this is the answeri, but in my class we only dealt with object on ramps and hills and cliffs.
449788
Doesn't matter whether it's dropped from a helicopter, skyscraper, or the goodyear blimp.  Terminal velocity is the velocity at which the air resistance of a falling object is equal to the acceleration due to to gravity, and the object can no longer can gain speed.  It's not a constant, as you get closer to the ground air pressure will increase, so the air's resistance will increase, and the teminal velocity will drop. The object will actually slow down.
724154
The answers describing the damped oscillations about the exact centre of mass of the planet are correct. The answer about terminal velocity meaning you don't go much beyond the centre is completely correct. You rock rock back and forth about the centre for a while, each time not getting as far away as the last. The distance would decrease proportionally to e^(at) where a is dependent on friction.\n\nBelieve me, I wish I didn't know this, been studying mechanics, including harmonic motion, for three days now for an exam!
950828
Gravity only "slows" the bullet during 1/2 of its total trip, during the other half, gravity accelerates the bullet.\n\nAs the bullet rises into the air its velocity is continually decreasing as the force of its own weight (force due to gravity) as well as the force of air resistance, act to accelerate it downward.  At some point the bullet will come to a stop and then begin falling.  Afterward, gravity acts to increase the bullet's speed while the air resistance works in the opposite direction.  The magnitude of the force of air resistance depends on the speed of the bullet (the faster it does, the harder the air pushes back on it).  Eventually, the bullet will reach what is known as the "terminal velocity", where the force of gravity equals the force of air resistance, and the bullet stops accelerating (net force = 0).\nThe terminal velocity for an object depends on several factors including the object's mass and shape, as well as the density of the fluid it is moving through.  The terminal velocity for a bullet is lower than the speed at which it comes out of the gun in most cases, so the bullet will never be traveling as fast as it was while leaving the barrel when it hits the ground...but still, I would still not want it to hit me.
694476
In your lab you are dealing with the relationship of:\nf=m*a\nso a=f/m\n\nIf you plot the values of acceleration versus mass holding force constant, this is a non-linear graph, so it will depend on the range of your data.  As m gets large, the slope will approach zero.  As m gets small the slope will approach infinity, which makes sense.  The instantaneous slope of this line is -f/m^2.  So your error will depend on the relative magnitude of the mass with respect to the net force.\nIf you plot the values of acceleration versus net force while holding the mass constant, the slope of this line is 1/m.  Your error will depend on the accuracy of the system.  If the coefficient of kinetic friction is not a constant, or the force applied is not a constant, then error will be introduced into the system.\n\nj
1003574
the dominoes will reach a terminal velocity.. based upon their weight and the friction of the air as they fall... just like dropping one from an airplane.. and having it always large face downward (not edgewise)
28151
Acceleration due to gravity (NOT change in acceleration), which could also be described as change in velocity.  That's assuming no air resistance.  If there is air resistance, they will eventually reach terminal velocity after which their velocity will be constant, but this is not implied by "free fall," and given that the question sounds like it's from a basic middle school or high school physics class I'm guessing you're supposed to ignore air resistance.
287006
Terminal Velocity.  It depends on the weight of the object because it is based on gravity.  If there is force applied to the object then it could travel faster for moments then gravity would slow it down.
338445
The net external force acting on a particle is equal to the product of the mass and its acceleration.\n\nF = ma\nwhere, F = the net external force\n       m = the mass of the particle\n       a = the acceleration of the particle
291004
if an object is floating, the buoyant force on it equals its weight -- the vertical forces have to sum to zero.  try drawing a free body diagram to convince yourself of this.
287902
Air resistance has little to do with terminal velocity. In fact the terminal velocity is lower on the moon than on earth, because the attraction of the moon for a body is less. The terminal velocity is equal to the impact velocity that a body would acquire if it fell unimpeded from a great distance away (distant space). That is the initial velocity an outbound object would need to totally escape from the moon or earth. Since the earth is more massive, its escape velocity would be greater, and the presence of air (drag) would be to make escape slightly more difficult from earth.
841193
It's the fall velocity at which air resistance exactly balances gravitational force, so the falling object no longer accelerates. However, it varies with air density, which changes as the falling object's altitude declines. Gravitational force also changes with altitude because it depends on the distance to the earth's center. As a result, a falling object is hardly ever going to fall at exactly its terminal velocity, since it has to "chase" the changing value of that velocity.
262699
If the object is dropped from a low enough altitude, it will impact before reaching terminal velocity, which is defined as "the speed at which a particular body's weight is balanced by its drag as it falls through a particular fluid" (ref.). Its maximum speed will be its speed at impact. Note that terminal velocity is a function of air density (which increases rapidly with decreasing altitude) and gravitational acceleration (which increases slowly with decreasing altitude).\nOtherwise the maximum speed is the terminal velocity at some point before impact and not the velocity at impact; in fact the peak speed may be much higher than either the velocity or the terminal velocity at impact. \nConsider an object dropped from a few hundred thousand feet altitude where the air is too thin to exert appreciable drag. Primarily due to gravity, the object reaches a fall rate of, say, a thousand ft/sec. As both fall rate and air density increase, eventually the drag force reaches the object's weight and it is at peak velocity and at terminal velocity for that altitude. After this, as air density increases further the terminal velocity decreases and the decelerating object's speed continually "seeks" the terminal velocity. Thus in this scenario the object reaches a peak speed that is greater than its final speed and its (decreasing) terminal velocity, which may be only a couple of hundred feet per second at impact.