Mechanical Energy & Work-Energy Theorem. Mechanical Energy Mechanical Energy consists of: Energy due to the relative position of the interacting objects.

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Presentation transcript:

Mechanical Energy & Work-Energy Theorem

Mechanical Energy Mechanical Energy consists of: Energy due to the relative position of the interacting objects (Gravitational Potential Energy, E g ) Energy due to the object's motion (Kinetic Energy, E k ) E mechanical = E g + E k

Review: Conservation of Energy Energy cannot be created or destroyed Energy can only be transferred from one form to another 2 nd Law of Thermodynamics:  During any energy transformation process, there will always be losses (usually in the form of thermal energy) The 2 nd law is put aside (for now) as we study the work-energy theorem

Work-Energy Theorem Work done on the object = Energy gained by object (Energy change of the object) More work had to be done to stop the car with the highest kinetic energy.

Deriving Gravitational Potential Energy An object of mass m is lifted with a constant force, F A. How much work is done on the object if it's lifted vertically to a height, h? FBD: F A F g F A = F g the work done on this object is the same as the potential energy gained

Potential Energy Examples

How much potential energy? A=30J C=20J B=30J D=10J E=0J (lost 30J) (Potential Energy is released) (In the form of Kinetic Energy, E k =30J) (All 3 have the same P.E.)

Deriving Kinetic Energy using Kinematics & Newton's laws A box is pushed from rest to a velocity of v. Calculate the work done on the box. What are we assuming? Applied force is in the same direction as displacement!! It’s also the only force acting on the box!

Deriving Kinetic Energy using kinematics & Newton's laws (continued) Assuming the box starts from rest, then u = 0 The work done (W) to move the box to a speed of v is equal to the kinetic energy (E k ) gained by the box. Total Mechanical Energy = Eg + Ek =

Energy Transfer Examples

Energy Transfer Problem 1 1 PE=0J KE=1920J ME=1920J 2 PE=588.6J KE=1331.4J ME=1920J V=6.66m/s 3 PE=1920 J KE=0J ME=1920J V=0m/s h=3.26m

Energy Transfer Problem 2 1 PE=1962J KE=0J ME=1962J v=0m/s 2 PE=1472J KE=490J ME=1962J v=4.43m/s 3 PE=0J KE=1962J ME=1962J v=8.86m/s 4 PE=1062J KE=900J ME=1962J h=2.17m

A Fun Energy Transfer Example

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