*Which unit of measuement do we use for both work and kinetic energy?\n\n*How can work transfer into kinetic energy?\n\n*A 2kg object is pushed with a force of 10 newtons and travels 2 meters, how much work was done? : The object had a velocity of 4 meters/second, what was it's kinetic energy?
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Energy Unit = Joule = 1 kg x m^2/s^2 = 1 N x m\n\nHow can work transfer into keinetic energy?\nThink in a pendulum in which these two are constantly tranfered into one another.\n\nWork = Force x Distance \n\nwork = 10N x 2m = 20 N x m\n\nKinetic energy = Mass x Velocity^2 / 2\n\nkinetic energy = 2kg x (4m/s)^2 / 2 = 16 N x m
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the units don't work out right for E=mc^3.\n\nc has units meters/second.\nm has units kilograms.\n\nE is energy, which has units of Joules, and 1 J = 1 kg * m^2 / s^2.\n\nwhen the equation is E=mc^2, the units work out correctly.  an equation where the units don't work out is nonsensical in physics.
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To lift an object from a ground we must apply a force equal to the weight of the object and move it to the required height.\n\nThus to do a work two things are required, a force and a distance. \n\nWork is the product of these two. \n\nIf a body of weight 32N is to be moved up through a distance of 1m we have to do a work of 32 joule totally. \n\nThat is we apply a force of 32N and move it through 1m.\n\nThe same work can be done by a boy who cannot exert a force more than 16 N.\n\nThis is possible because, the boy has to move 16N force through a distance of 2m instead of usual 1m.\n\nWe can apply 1N force and move it through 32m. \n\nOnly thing is that the product of force and distance must be 32 joule.\n\nAll simple machines are based on this principle. \n\nWhen we reduce the force the distance is increased; when we increase the force the distance is reduced. But the product remains the same. That is the total work done is the same.
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The work done on the car by the net force equals the change on the car's kinetic energy.\n\nWork = Force * distance\nKE = 1/2 mv^2\n\nWe are given a mass for the car and the final velocity, so from this we can calculate the car's final KE,\nKE = 1/2 (2000 kg) * (2 m/s)^2\nKE = 4000 Joules\n\nSince in this case, KE = W\nW = 4000 Joules = F * d\n\nWe know the net force to be the sum of all the forces acting on th car.  Since there is a "forward" force of 1140 Newtons and a "backward" (what we will call 'negative') force acting on the car of 950 N, the net force is 1140 - 950 = 190 Newtons\n\nW = 4000 J = (190 N) * d\nd = 4000 J / 190 N\nd = 21.05 meters
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kg m^2\n------- = Joule\ns^2\n\nb = kg m/s
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a. The spring constant for the inner spring can be determined by the fact that the force it exerts goes from 0 to 20 units (I can't read what the units are. Are they Newtons?) over a difference of 0.10 m. So its spring constant is 200 force units per meter, or 2 force units per centimeter.\n\nb. From point A to point B, the average force exerted by the spring is 10 units. Because the force changes linearly over the distance of 0.10 m, we can use this average and calculate the work done on the 5 kilogram object as 10 x 0.10 = 1.0 Force unit-meters.\n\nc. From point B to point C, the average force is 40 units, and the distance is 0.05 m, so the work done is 40 x 0.05 = 2.0 Force unit-meters. (Again, the force changes linearly over the 0.05 meters.\n\nd. The initial velocity of the object will be such that its kinetic energy equals the amount of work done by the two springs to slow its velocity to zero (i.e., to absorb all of its kinetic energy).\nJust use the formula for kinetic energy and set it equal to 3.0 force unit-meters.\n\ne. To do this step, you have to recognize that the inner spring exerts a force of 20 units at point B, and (assuming its spring constant does not change) it will exert a force of 30 units at point C. So the outer spring, which provides the rest of the force, is providing 0 units at point B and 30 units at point C, a change of 30 units in 0.05 meters, or 600 units per meter (= 6 units per centimeter).
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WORK IS EQUAL TO ZERO (READ ALL FOR EXPLAINATION)\nmass of body=30kg\nits weight=30kg X 9.8m/s^2\nweight= 294N\n\nBUT HERE WE DON'T HAVE TO DEAL WITH ITS WEIGHT.\n\nAs the body is moving at constant speed, it means that it's not accelerating and thus no force is being applied to it.\n\nOr in other way\n\nthe frictional force which is 20N is canceling the force with which the boy is moving the body. Thus the boy is also exerting a 20N force (20N - x = 0, of course x=20N)\n\nTHE WORK IS ZERO BECAUSE THE FORCE IS ZERO TOO.\n\nsince the frictional force is cancelling the force of the boy.\n\n20N(frictional force)-20N(force applied)=0\nResultant force = 0\n\nFIND WORK NOW\n\nwork =  net force X distance\nwork =  0  X  5m\nwork =  0 joule\nNo work is done\n\nLISTEN\nThe person below me by the name of sblogs is incorrect as I think. Let me prove this by giving some EXAMPLES\nPUSH THE WALL BUT THE WALL WILL NOT MOVE FROM ITS PLACE THOUGH YOU ARE WASTING ENERGY BUT THE WORK DONE IS ZERO.\nBECAUSE THE WALL DOESN'T MOVE THROUGH SOME DISTANCE.\nWORK= FORCE X DISTANCE\nWORK= (FORCE APPLIED ON THE WALL) X (ZERO METERS)\nWORK= F X 0\nWORK= 0 JOULE
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1.force*distanc=300*10\n=3000 joules\n\n2.force=work/distance\n=5000/100=50N\n\n3.work/force=distance\n1600/80=20m\n\n4.D=12000J\n\n5 The correct unit for work is:\nB Joules
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work was defined as a force acting upon an object to cause a displacement. When a force acts to cause an object to be displaced, three quantities must be known in order to calculate the work. Those three quantities are force, displacement and the angle between the force and the displacement. The work is subsequently calculated as force*displacement*cosine(theta) where theta is the angle between the force and the displacement vectors. In this part of Lesson 1, the concepts and mathematics of work will be applied in order to analyze a variety of physical situations.\n\n\nw = 100N *10m* cos(0 degrees0 = 1000J\n\nSince F and d are in the same direction,the angle is 0 degrees.Forces do work when there is some component of force in the same or opposite direction of the displacement.
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The last poster needs to have his dynamic physics retaught, or stop trying to sound like an engineer.\n\nThe first mistake, is identifying newtons = force x distance.  This is actually a measure of work or energy and measured in joules.  The second mistake identifying force as being measured in joules.  Force is measured in Newtons.  Since it was Isaac Newton in the second law of motion that defined Force.\n\nYour question cannot really be answered without more information.  What is the direction of motion, horizontally or vertically?  \n\nVerically?  Then 150 lbs of force.  Pounds were once used to measure force.  If you want to use metric, convert the 150 lbs to kilograms, multiply by 9.8 and then you must exert at least this much force in Newtons to overcome the force of gravity.  150 lbs is 68.04 kg.\n\nHorizontally, what type of surface?  Determine the coefficient of friction, and multiply it by the mass in kilograms.  Then this is the amount of force you need to exert to overcome the force of friction.  There are two types of friction, static and dynamic.  In other words it takes more force to move an object at rest, but once the object is in motion it takes less force to keep it move (opposing the dynamic force of friction).\n\nWatts by the way is a measurement of power, which is energy derived by time, Power = dE/dt.  Joules per second.
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Mass is 3500 kg.  Distance moved is 200 meters (up).\n\nForce is usually calculated as mass times acceleration.  The SI unit used to measure force is the newton (symbol N), which is equivalent to kg*m/s^2.  The acceleration here is the acceleration due to gravity or\n\ng = 9.80665 m/s2\n\nThe force = m*g = 3500 kg * 9.81 m/s^2 = 34335 Newtons.\n\nThe energy (or work done) is the force times the height or\n\n34335 Newtons * 200 meters = 686700 joules.\n\nTo calculate the power required, you would have to know how long the movement took in seconds.
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WORK = force x distance  So you would need something the work would be done on. you would need to exert a force on it and someplace for it to go.
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the kinetic energy of the sled and person is equal to the work done by the friction to stop the sled.  Since work is force times distance in this case, the force can be calculated.  From the force you can find the coefficient from the definition of friction by dividing out the normal force.  \n\nor literally, 0.5*m*v^2 = F*d = mu*m*g. solving for mu\nmu= v^2 / (2*d*g) = 0.016\n\ni think....
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what your calling energy is really Net Force.  or Mass x Acceleration.  \n\nin the first example, the energy wasn't gone, if you assume that the people have the same mass and were traveling at teh same velocities when they reached teh end then the net force acting on the system is 0 because the force (ma) of the first person is equeal to the force of the second person, so they cancel.  \n\nin the second example, a centrepedial force, or force towards teh center, is being applied from teh rope on the people holding it.  this causes the direction of their velocity to change.  using dot products we can find teh work done, which is change in kinetic energy.  you will find that the work done on one person is equal and opposite to the work done on the other.
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51kg x 25 m = energy used\n\ndivide the energy used by 0.30 to get the second answer
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You don't need to find the mass if use conservation of energy to solve the problem.\n\nThe kinetic energy of the can initially is:\n1/2*m*v^2\n=.5*m*2^2\n=2*m\nand when it comes to rest it is zero\nThe energy was lost to the work done by friction.\n\nthe force of friction times displacement is the work:\n\nu*g*m*d\n=.2*9.8*m*d\n=19.6*m*d\nnow set the energy of the system equal to zero:\n2*m-19.6*m*d=0\ndivide out the mass and solve for d:\nd=19.6/2\n=9.8m\n\nj
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Here given data is\nMass = 2 * 10^3 kg\nKinetic Energy = 4 * 10^5 Joules\n\nThe formula for kinetic energy is \nKinetic Energy  = 0.5 * m * v^2 where\nm - mass in kg\nv - velocity in m/s\n\nSo 4*10^5 = 0.5 * (2*10^3) * v^2\nThus solving we get v = 20 m/s           \n\nHope u got wat u wanted........