1. Energy Transformation

2. PE vs. KE  We saw this on a quiz

3. Potential energy of lifted object  The work to lift an object to a height “h” is defined by W = Fd. Hence…  W = weight * h and…  Weight = F w = mg. Hence the PE of a raised object due to its height is…  PE = mgh  If the man has a mass of 50 kg, and the height of the pole is 20 meters, what is the PE?  PE = (50kg)(10m/s 2 )(20m) = 10000 J

4. Work transfers energy to an object  Describe how the energy changes in each of the cases above.  Did it make a difference how the work was done?

5. Is energy being changed?  Describe how the energy changes in each of the cases above.  We’ve done this before.

6. Energy transformations  Is work being done?  How do you know?

7. Is work being done?  Remember, W = Fd

8. Potential energy of lifted object  Obviously the rock had stored energy at the top of the pinnacle, because when it began to roll down the hill it picks up speed – kinetic energy.  The potential energy (PE) is based on the weight of the object (mg) and the height. (PE = mgh)  The kinetic energy (KE) is based on the speed. (KE = ½ mv 2 )

9. Potential energy of lifted object  Describe how the energy changes during a roller coaster ride.

10. Potential energy of a spring  Is there potential energy stored in the spring at the left?  Once compressed, there is stored energy.  The energy is dependent on the amount of compression or stretching of the spring.

11. Potential energy of a spring  The amount of energy stored when a spring is compressed or stretched depends on the stiffness of the spring material.  F = -kx where k is a spring constant based on the spring.  If the spring is compressed twice as far, two times the force was required to stretch it to that distance.

12. Potential energy of a spring  The spring shown is stretched 0.30m.  As a result, it has stored PE.  Using PE = ½ kx 2 and if k = 5.0 N/m, what work was done to stretch it the 0.30m?  W = ½ kx 2 = (½)(5.0N/m)(0.30m) 2 = 0.225 J.  What average force was applied?  W = Fd  F = 0.225J/0.3m = 0.75N  However, F = kx = (5.0N/m)(0.3m) = 1.5N tells us that the force of the stretched spring is now 1.5N. Problem?  Since the force when it isn’t stretched is 0.0 N, the avg. F = (0 + 1.5)/2 = 0.75N.

13. Potential energy of a spring  The slope of the line above is…  That’s right  k  The area under the curve = ½ (kx)(x) = ½ kx 2 Work is force integrated with respect to distance.