Hey. All well and good. That’s a reallyimportant result, so we’ll revisit it here. Mass m, initially atx0, with velocity v0. Constant total forceF in the x guidance accelerates it ata equals F over m. Constant force givesconstant acceleration. So we can use equation threefrom kinematics to pertain the displacement– x minus x0– to the initial andfinal velocities. The total impel is constant andparallel to the displacement, so the work is Ftimes x minus x0. Substituting for Newton’ssecond constitution, we have work equals m a hours x minus x0. But, retaining equationthree, let’s multiply by 1 and write the 1 as 1/2 times 2. Now we can substitutefor 2a x minus x0. That’s just v squaredminus v0 squared. So now we have totalwork done equals 1/2 mv squared minus 1/2 mv0 squared. This is the change in thequantity 1/2 mv squared. That is such an important resultthat we make that sum a identify. We call that quantitythe kinetic energy. Kinetic energy is definedas 1/2 mv squared. What we’ve just proved isthat the total work done by all the forcesacting on an object is equal to the changein its kinetic energy. Actually, we’ve only shownthis for constant force in one dimension. But it’s true forvariable personnels as well. And it’s called thework-energy theorem. Again, let’s check the units. mv squared is kilogrammes timesmetres per second squared. Take one of the metres outsideand use Newton’s second regulation to remind us that kilogrammemetres per second squared is force– units, Newtons. And Newtons timesmetres equals joules. Let’s do a question toget used to the units ..
