1. Aim: How do we solve conservation of energy problems?

2. Mechanical Energy
The mechanical energy of an object is the sum of its kinetic energy and potential energy. In the absence of external forces, the mechanical energy of an object does not change.

3. Example 1-Fill in Table Assume vi = 0 m/s at point 1

4. At position 1
At position 1, PE=mgh=50kg(9.8m/s2)(4m) PE=1960 J KE=1/2mv2=1/2(50kg)(0m/s)2=0J E=PE+KE=1960J=1920 J

5. At position 2
E=1960 J PE=mgh=50kg (9.8m/s2)(3m)=1470 J E=KE+PE 1960 J=KE+1470 J KE=490 J KE=1/2mv2 490 J=1/2(50kg)v2 v=4.42 m/s

6. At position 3
E=1960 J PE=mgh=50kg(9.8m/s2)(0m)=0J E=PE+KE 1960 J= 0 J+KE KE=1960 J KE=1/2mv J=1/2(50kg)v2 v=8.9 m/d

7. At position 4
E=1960 J KE=1/2mv2=1/2(50kg)(6m/s)2=900 J E=PE+KE 1960 J=PE+900 J PE=1060 J PE=mgh 1060 J=50kg(9.8m/s2)h h=2.2 m

8. Example 2 A 2kg ball is released from rest at point A.
Calculate the PE at point A.
PE=mgh=2kg(9.8m/s2)(40m)=784 J
2) Calculate the speed of mass at point E.
The potential energy at A is equal to the kinetic energy at E.
KE=1/2mv2
784 J = ½(2kg)v v=28 m/s
3) What will be the maximum height that the 2 kg ball will reach?
The maximum height is 40 m

9. Pendulum System
At the lowest point of its swing, the pendulum is 0.3 m below its release point at point A. The mass of the pendulum bob is 2 kg.
Calculate the speed of the pendulum at point B.
The potential energy at A is equal to the kinetic energy at B.
PE=mgh=2kg(9.8m/s2)(.3m)=5.88 J
KE=1/2mv2
5.88 J =1/2(2kg)v v=2.4 m/s

10. Example 4- High Jumper Problem
Calculate the kinetic energy and the velocity required for a 65 kg pole vaulter to pass over a 4.5 m high bar. Assume the vaulter’s center of mass is initially 0.90 m off the ground and reaches its maximum height at the level of the bar itself.

11. The kinetic energy at the bottom is equal to the potential energy at the top. PE=mgh=65 kg (9.8m/s2)(4.5 m-0.9m) PE= J KE=1/2 mv J=1/2(65 kg)v2 v=8.4 m/s