differentiate the net force acting on a body between motion in a straight line and along a curved path?
83052
Forces occur in pairs. Newtons third law.\nIf this is the case the forces exist as equal and opposite.\n [Fo ]=[-Fi]\nNet force = Fo minus Fi= zero!\n  Moving forces can be directional and vectorial.\nAlso they change continueously during the motion till they Atenuate at the end of their cycle.To identify the force you have to identify the power that has cause the motion of the mass and the direction of interaction. The power causing the motion is always equal to the magnetude of the force multiplied by the the speed that the force moves.\n\nCurvilenear motion indicates that the power of two moving masses interact and cause the two masses to move around a common pivot point. Thus the masses are locked together into an orbit forming a composite mass structure.In that case the moving masses experience a curved path.\nThan you have to understand what is  a circle. A circle consists of an infinite integral of short straight lines, and each straight line having different direction.\n\nDont feel bad if you cant fully understand why. . I dont find it easy myself.
896953
The net force is the difference between the two weights.\n\nHowever, there is not enough information in the question to answer without making some assumptions.\n\nYou are not given the mass of the man and/or the bricks. One can calculate the mass if given the force of gravity, however you aren't given that either.\n\nOne method of solving is:\n\nASSUMING that this system is on the surface of the Earth, you can find the two masses, using F=mg with g=9.8N/kg. Once you've got that you can find the acceleration using F=ma since you know the net force and the mass of the system.\n\nMake sure you indicate which direction the acceleration is in (in this case, which object is falling, and which is being raised.
256104
Read Newton's 3rd law of motion.\nAlso, use F=ma to figure out your acceleration.
716944
If there is no net force (i.e. sum of forces is zero) then there will be no change in velocity. So if the object was going 15 m/s and the net force is zero, then it will still be going 15 m/, (for 2 sec = 30 meters traveled). Same concept for your jet car - constant velocity means no net force ( there may be forces acting on it, but if the velocity is constant, they must sum to zero - net force is zero).\n\nAs Newton's first law says, "An object in motion tends to stay in motion unless acted on by an outside force." So constant velocity (i.e. stays in motion at the same velocity) means no net force acting on the object.
774794
Remember the net work is the change in Kinetic energy.  In order for a force to do work, it must be parallel the motion.\n\nW = f (dot) d   =  f*d*cos(theta)\n\n Also, Mechanical (and Total) energy is conserved. Thus, any increase in kinetic energy is coupled with an equal decrease in potential energy. \n\n/&#92;K = -/&#92;U = Wnet
210676
Use free body diagram to balance the forces and u can derive the equation of motion \nIf u want i can give u full derivation.\n"::
755944
Net force is required to change the motion of an object, or accelerate the object. Often forces are balanced, so net force = 0, so motion does not change.\n\nWork is in the same units as energy, not force. Work is related to force by \nW = F*d*cos(theta) where F is the applied force, d is the distance through which the force has been applied, and theta is the angle (if any) between the force and the distance. \n\nFor example if a 100 Newton vertical force is applied to lift an object to a table 5 meters above the starting point, 500 Newton*meters, or 500 joules, of work have been done.
9163
There is a force, called centripetal force, that exerts on every body in rotational movement. If that force would not exist the body would go in straight line.\nThe magnitude of that force is\n\nF=m ω^2 r\n\nwhere m [kg] is the mass of the body, ω [rad/s] the angular velocity of the body and r [m] the radius of the trajectory.\n\nWe have the relation between rpm and angular velocity:\n\nrpm=60ω/2π => ω=πrpm/30\n\nHence, F=m π^2 rpm^2 r/900\n\nAccording to Newton: F=m a  (where a is the acceleration in m/s^2)\n\nThus a=π^2 rpm^2 r/900\n\nto get the expression in g you divide both sides of the equation by gravity's acceleration g (~9,81 m/s^2)\n\nSo RCF=(π^2 rpm^2 r) / (900 g) \n(RCF=relative centrifugal force)\nRCF=1.118*(10^-3) * rpm^2 * r        in SI units
1164784
Add up one opposite vector (going up). In physics, they call it NORMAL FORCE.
1062743
D. The road.\n\nIf you draw a free-body diagram you'll see a frictional force vector pointing in the direction of motion.
863293
In the opposite direction of the applied force or motion
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).
193077
Force=Mass x Acceleration\n\nSo Acceleration = Force/Mass\n\nHope this helps
966154
The magnitude of a force is measured in newtons(N)
979740
During the first 10 m the force increases in a linear fashion. Because it's linear, you can average it for the first 10m: 5 N. This easy way will give the same velocity at the end of the 10 m as if you did it by area under the curve. According to Newton's 2nd law, this gives the object acceleration a1:\nF = m*a1 where F = 5 N and m = 6.90 kg.\n\nUse a1 in the equation V^2 = Vo^2 + 2*a1*10 m where Vo = 0. Solve for V.\n\nFor the second 10 m section, the force is steady at 10 N. Again use Newton's 2nd law to get a2:\nF = m*a2 where F = 10 N and m = 6.90 kg.\n\nUse a2 in the equation V^2 = Vo^2 + 2*a2*10 m where Vo = V from the last part.  Solve for V.
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.
903675
The frictional force acting against the motion is F\n\n= (mu)x mg x cos a. \n\n= 0.2 x 22 x 9.8 x cos 25 = 39 N. \n\nThe net force acting along the inclined plane\n\n= (mg sin a +39) Since both the forces are acting opposite to the motion.\n\n= 22 x 9.8 x sin 25 + 39\n\n= 130. 12 N. \n\nIf x is the distance moved along the plane, the work done by this net force is Fx.\n\nThis work done is used to remove the kinetic energy of the body = 0.5 m v v.\n\nFx = 0.5 m v v \n\nx = 0.5 x 22 x 3.9 x 3.9 / 130.12 = 1.28m.