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Current time:0:00Total duration:5:48

AP.PHYS:

INT‑3.E (EU)

, INT‑3.E.1 (EK)

, INT‑3.E.1.1 (LO)

, INT‑3.E.1.4 (LO)

in order to transfer energy to an object you've got to exert a force on that object the amount of energy transferred by a force is called the work done by that force the formula to find the work done by a particular force on an object is W equals FD cosine theta W refers to the work done by the force F in other words W is telling you the amount of energy that the force F is giving to the object F refers to the size of the particular force doing the work D is the displacement of the object how far it moved while the force was exerted on it and the theta in cosine theta refers to the angle between the force doing the work and the displacement of the object you might be wondering what this cosine theta is doing in here this cosine theta is in this formula cuz the only part of the force that does work is the component that lies along the direction of the displacement the component of the force that lies perpendicular to the direction of motion doesn't actually do any work we notice a few things about this formula the units for work are Newton's times meters which we call joules joules are the same unit that we measure energy in which makes sense because work is telling you the amount of joules given to or taken away from an object or a system if the value of the work done comes out to be positive for a particular force it means that that force is trying to give the object energy the work done by a force will be positive if that force or a component of that force points in the same direction as the displacement and if the value of the work done comes out to be negative it means that that force is trying to take away energy from the object the work done by a force will be negative if that force or a component of that force points in the opposite direction as the displacement if a force points in a direction that's perpendicular to the displacement the work done by that force is zero which means it's neither giving nor taking away energy from that object another way that the work via force could be zero is if the object doesn't move since the displacement would be zero so the force you exert by holding a very heavy weight above your head does not do any work on the weight since the weight is not moving so this formula represents the definition of the work done by a particular force but what if we wanted to know the network or total work done on an object we could just find the individual amounts of work done by each particular force and add them up but there's actually a trick to figuring out the net work done on an object to keep things simple let's assume that all the forces already lie along the direction of the displacement that way we can get rid of the cosine theta term since we're talking about the net work done on an object I'm gonna replace F with the net force on that object now we know that the net force is always equal to the mass times the acceleration so we can replace F net with M times a so we find that the net work is equal to the mass times the acceleration times the displacement I want to write this equation in terms of the velocities and not the acceleration times the displacement so I'm going to ask that you recall a 1d kinematics equation that looked like this the final velocity squared equals the initial velocity squared plus two times the acceleration times the displacement in order to use this kinematic formula we've got to assume that the acceleration is constant which means we're assuming that the net force on this object is constant even though it seems like we're making a lot of assumptions here getting rid of the cosine theta and assuming the forces are constant none of those assumptions are actually required to derive the result we're going to attain they just make this derivation a lot simpler so looking at this kinematic formula we see that it also has acceleration times displacement so I'm just gonna isolate the acceleration times the displacement on one side of the equation and I get that a times D equals V final squared minus V initial squared divided by two since this is what a times D equals I can replace the a times D in my network formula and I find that the network is equal to the mass times the quantity V final squared minus V initial squared divided by two if I multiply the terms in this expression I get that the network is equal to one-half mass times the final velocity squared minus one-half mass times the initial velocity squared in other words the network or total work is equal to the difference between the final and initial values of one-half MV squared this quantity one-half M times V squared is what we call the kinetic energy of the object so you'll often hear that the net work done on an object is equal to the change in the kinetic energy of that object and this expression is often called the work-energy principle since it relates the net work done on an object to the kinetic energy gained or lost by that object if the net work done is positive the kinetic energy is going to increase and the object is going to speed up if the net work done on an object is negative the kinetic energy of that objects going to decrease which means it's going to slow down and if the net work done on an object is zero it means the kinetic energy of that object is going to stay the same which means the object maintains a constant speed you