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WORK ENERGY THEOREM According to the work-energy theorem, the net work on an object causes a change in the kinetic energy of the object. The formula for net work is net work = change in kinetic energy = final kinetic energy - initial kinetic energy. The work W done by the net force on a particle equals the change in the particle's kinetic energy KE: W=ΔKE= m The work-energy theorem can be derived from Newton's second law. Work transfers energy from one place to another or one form to another. In more general systems than the particle system mentioned here, work can change the potential energy of a mechanical device, the heat energy in a thermal system, or the electrical energy in an electrical device. EK= PE=mgh The unit for work and energy is J ME=mgh+ h= ; v= 2 ℎ − Examples: 1. Calculate the work needed to move an object in a distance of 8.3m using a force of 250N. W=Fd=8.3m ∙ 250N=2075J 2. Find the distance travelled by an object with mass 45 kg moving at 22m/s acted on by a 100N force. F∙d= m 100d= (45)(484) d= =108.9m 3. Find the velocity at which a 55kg object is travelling when a 70N force acts on it over a distance of 30m. F∙d= m 70∙30= (55)(v2) v2= . v=8.7m/s 4. Find the potential energy for a 27 kg object raised to a height of 15m. PE=mgh PE=27∙9.8∙15 PE=3969J 5. Find the velocity of an object just before it hits the ground, if it is dropped from a height of 23m. v= 2 ℎ= 2(9.8)(23)=21.23m/s 6. Find the height from which an object falls if just before it hits the ground it has a velocity of 28m/s. h= = =40m . 7. If an object hits the ground with a velocity of 16m/s, at what height did it have a velocity of 8m/s. ME1=ME2 mgh1+ = mgh2+ v1=8m/s 2 h=? 2 m(gh1+ v1 )=m(gh2+ v2 ) gh1+ v12=gh2+ v22 h2is 0 because the object hits the ground. v2=16m/s gh1= h1= = ∙ . =9.8m 8. Find the original height from which an object is dropped, if at a height of 40m it is moving at 20m/s. gh1=gh2+ v2 2 gh1=(9.8)(40)+ (20) h1= . ( ) . h1=? h2=40m v2=20m/s =60.4m