how can work be done when the net force equals zero? Explain?
960318
Remember that there is friction acting on the car from the movement of the wheels over the road. Work must therefore be done to overcome this friction.
905447
The net force is 0 Newtons. The force of friction is 100N.\n\nThe key here is that\n1. No acceleration.\n2. Horizontal force only.\n\nThis means that the forces are balanced in the horizontal direction. The object isn't accelerating so net F = ma = 0.
181079
Work equals Force * Distance,\nW = F*d\n\nSo you could say that work was proportional to the force applied or to the distance over which the work was done.\n\nWork also equals a change in energy, whatever energy that may be...kinetic energy, potential energy, what-have-you.\n\nIf some power is being exerted, work would be proportional to the time the power was being applied,\nW = P*t
260289
Stephen_mgi is correct.\n\nForce x distance moved is the work done.\n\n Suppose a 10 N force is moved through a distance 5m the work done is 50 joule.\n\nThe same work can be done in many number of ways; for example  5 N force when moves through 10 m or 1N force moves through 50 m; or 25N force moves through 2m etc.\n\nThus when force is reduced distance is increased. Any one will feel it convenience to reduce the force rather than to give huge force and jump up to go to a higher place.\n\nIn the stair case when we increase the distance by walking in a diagonal way we reduce the force and feel it easy.\n\nThe stair case is a simple machine which is designed at a slope. \n\nBy walking diagonally we still reduce the slope to our convenience.
46266
When it is said that that the net force is zero, it means yes, there is gravitational forces acting at the cenetr of the earth, but being at the center, with equal amounts of earth mass in every direction, the forces pulling on you at the center would all cancel each other out, resulting in the <perception> of having no forces acting on you, when in reality, gravity is always at work.
1147561
Net force on object = Force applied - Friction force\n\nSince velocity is constant, the acceleration of the object is zero\n\nSo, Net force = mass * acceleration = zero\n\n0 = Force applied - Friction force\nor\nForce applied = friction force \n= coefficient of friction * Normal force\n=.250 * 11.9 kg * 9.8 m/s^2\n=29.2 kg-m/s^2\n\nBox is decelerating at .250*9.8m/s^2\n\nand\n\ndistance traveled = (final velocity^2 - initial velocity^2)/2a\n=(0 - 3.2 m/s)  /  2*(-.250*9.8m/s^2)\n=.65 m
239996
Work done = force x distance moved by object i thinkin a straight line ,so ithink some conversion maybe needed.
846783
Mechanical work is a force applied through a distance, defined mathematically as the line integral of a scalar product of force and displacement vectors. Work is a scalar quantity which can be positive or negative.\n\nAs seen from the above definition, force can do positive, negative, or zero work. For instance, a centripetal force in uniform circular motion does zero work (because the scalar product of force and displacement vector is zero as they are orthogonal to each other). Another example is Lorentz magnetic force on moving electric charge which always does zero work because it is always orthogonal to the direction of motion of the charge.
936812
Uhhh..\nWork = Force * Distance..\nEnergy = Work
922769
Okay so start off by looking at the definition of work.\n\n"Work is a force exerted on an object that causes it to have a displacement in the same direction.\n\nW = Fd\n\nW = work (Joules), F = force (Newtons), d = displacement (metres)\n\nBy definition, 1 J of work is done by applying 1 N of force to move an object 1 m.\n\n    * Work has no direction, so it is a scalar quantity.\n    * Also, the direction of motion must be in the same direction (parallel) as the direction of the force, otherwise no work is being done."1\n\nSo there are two things you could look at here.\n\nW = Fd\n\nHow much displacement is there in one full cycle of something moving in a circle?  Displacement is the change in position so look at the position at the beginning and ending of the cycle.  There is no change in position so d=0 which gives\n\nW=F(0)\nW=0.\n\nNo work done.\n\nAnother thing you have to look at is "Also, the direction of motion must be in the same direction (parallel) as the direction of the force, otherwise no work is being done"\n\nIn uniform circular motion the movement of the object at any point in time is tangential to the circle while the force is always directly towards the centre of the circle so the movement and force are never in the same direction.  The definition of work demands that they be in the same direction for any work to have been done and, since they are not, no work has been done.\n\nShort answer: c. 0 J
912606
The simple answer is that the vector sum of the forces acting on an object must be zero if the object is to remain in "equilibrium" or in other words to have constant velocity.
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
777421
With zero net force there are so many examples. Book resting on the table. Vehicle moving with constant velocity. Rain drops falling with terminal velocity. Object floating on water. Mass hanging from a spring balance. \n\nBut the example of a known total force F, acting on mass m and producing accelerartion a = F/m in the mass m is difficult to give. This is so because of two reasons. Firstly because rarely only wellknown forces act on an object, most of the time not so                                   wellknown forces such as that of reaction tension and friction also act on the objcet. Secondly, strictly speaking there exists no definition of mass other than Newton's second law. Newton's Second law define both force and mass.  In this sense this is not a law but a definition of some sort. on the basis of which you can derive certain results of experiment which can be verified.
853308
The net force on the man is the sum of the force of gravity and the force on the ground. You've already calculated the net force, so simply subtract the man's weight from it. Remember that gravity is pulling him down, so you'll be subtracting a negative number (i.e. the final result should be greater, not less, than the net force).
834482
The answer is C. Although B seems to be a possible answer at first, there could still be an n-number of forces acting on the object such that the sum of all the force vectors is zero, thus making C the more complete answer. One of Newton's laws says that F=m*a (force on an object equals its mass times its acceleration), or a force of 1 Newton on a 1kg object will make it accelerate at a rate of 1m/s^2 in the same direction of the force.
997152
You mean Newton's 2nd law of motion, don’t you?\n\nForce = mass * acceleration\n\nI like this law, a lot, actually.\nIt may not seem very intuitive at first to someone trying to learn it, but with practice, you'll come to understand.\n\nThe basic concept is that a [net] force applied to an object causes it to accelerate.  When you push on something, it moves.  They key work being a "net" force.  If there is no net force (i.e. F = 0), there is no acceleration.\nThe force the chair is exerting on you, called the "Normal" force is necessary because while you are sitting in the chair are you falling towards the ground?  No, of course not.\n\nWe all know that Earth (gravity) exerts a force on you in the downward direction, called "Weight".  Without a balancing force, you would continue to accelerate downward.  If the chair did not exert a normal force back upward to cancel weight, there would be a net force acting upon you in the downward direction which would cause you to accelerate downward.\n\nA similar case is true for other objects, like when you push on a wall.  In this case, it is not gravity which exerts the force, you are doing it.  In order to keep the wall from accelerating when you push on it, the wall must exert a force back which is "normal" to its surface.  Eventually, however, the wall will not be able to provide ample normal force to counter your push and parts of the wall will begin to accelerate as the wall breaks.
21035
ya torque=f*d and since the distance is zero because the applied force is at the moment of inertia... Troque is zero! bahahaah oy i hate statics.. or physics.. same thing