Physical science support documents : 2004 curriculum  Page 54 
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Physical Science Support Document The Conservation of Mechanical Energy: Background Information 2. Calculate the average velocity for the falling object during the fall. ( average velocity of the system) Average velocity ( vav) = distance through which mass falls / average time 3. Calculate the Final Velocity of the falling object as it just reaches the floor. Use equation: Average velocity ( vav) = ½ ( vi + vf): assume that the acceleration was uniform. Therefore: vf = 2vav where vi = 0.0 m/ s This equation is no longer included in the NCSCOS but is useful here! 4. Calculate the potential energy of the hanging mass while suspended at its highest point. Use equation: PE = mgh where m= mass of hanging object in kilograms g = 9.8 m/ s2 h = height of hanging mass above floor in meters Units are important. All masses should be in kilograms and distances in meters. The unit of PE is Joules 5. What is the potential energy of the mass while on the floor? Use equation: PE = mgh where h = 0.0 meters ( assuming floor is assigned the ground level) Therefore: PE = 0.0 Joules 6. What was the change in potential energy as the object went from its highest position to its lowest position? ΔPE = PEf – PEi ( answer from number 5 – answer from number 4) This will give a negative value for potential energy. The interpretation is that PE decreases 7. Was the potential energy of the hanging mass lost? If not, what happened to it? The potential energy wasn’t lost; it was converted to kinetic energy. As potential energy decreased, the response of the system was to accelerate increased KE. 8. What was the Kinetic energy of the SYSTEM when the hanging mass was suspended at the pulley? Use equation: KE = 1/ 2 mv2 Where m is the mass of the system in kilograms and v is the velocity while the system is at rest ( 0.0m/ s) This gives a KE of 0.0 m/ s leading to a KE = 0.0 Joules 9. What was the kinetic energy of the SYSTEM when the hanging mass was just reaching the floor? Use the equation: KE = 1/ 2mv2, where m = mass of system ( kg) and v = velocity of the system when the hanging mass hits the floor. This value should be equal to vf found above. This value should be in m/ s. The actual student answer depends on the specific experiment. It is important to use the mass of the system here since the entire system is accelerating, not just the hanging mass. 51 June 2007
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Title  Physical science support documents : 2004 curriculum  Page 54 
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Full Text  Physical Science Support Document The Conservation of Mechanical Energy: Background Information 2. Calculate the average velocity for the falling object during the fall. ( average velocity of the system) Average velocity ( vav) = distance through which mass falls / average time 3. Calculate the Final Velocity of the falling object as it just reaches the floor. Use equation: Average velocity ( vav) = ½ ( vi + vf): assume that the acceleration was uniform. Therefore: vf = 2vav where vi = 0.0 m/ s This equation is no longer included in the NCSCOS but is useful here! 4. Calculate the potential energy of the hanging mass while suspended at its highest point. Use equation: PE = mgh where m= mass of hanging object in kilograms g = 9.8 m/ s2 h = height of hanging mass above floor in meters Units are important. All masses should be in kilograms and distances in meters. The unit of PE is Joules 5. What is the potential energy of the mass while on the floor? Use equation: PE = mgh where h = 0.0 meters ( assuming floor is assigned the ground level) Therefore: PE = 0.0 Joules 6. What was the change in potential energy as the object went from its highest position to its lowest position? ΔPE = PEf – PEi ( answer from number 5 – answer from number 4) This will give a negative value for potential energy. The interpretation is that PE decreases 7. Was the potential energy of the hanging mass lost? If not, what happened to it? The potential energy wasn’t lost; it was converted to kinetic energy. As potential energy decreased, the response of the system was to accelerate increased KE. 8. What was the Kinetic energy of the SYSTEM when the hanging mass was suspended at the pulley? Use equation: KE = 1/ 2 mv2 Where m is the mass of the system in kilograms and v is the velocity while the system is at rest ( 0.0m/ s) This gives a KE of 0.0 m/ s leading to a KE = 0.0 Joules 9. What was the kinetic energy of the SYSTEM when the hanging mass was just reaching the floor? Use the equation: KE = 1/ 2mv2, where m = mass of system ( kg) and v = velocity of the system when the hanging mass hits the floor. This value should be equal to vf found above. This value should be in m/ s. The actual student answer depends on the specific experiment. It is important to use the mass of the system here since the entire system is accelerating, not just the hanging mass. 51 June 2007 
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