WORK-ENERGY PRINCIPLE: The work of a resultant external force on a body is equal to the change in kinetic energy of the body. W =  K Units: Joules (J)

Work Energy Theorem – Proof! A net force acts on a moving object vo v Kinetic Energy F m x

What average force F is necessary to stop a 16 g bullet traveling at 260 m/s as it penetrates into wood at a distance of 12 cm? W = ΔK m = 0.016 kg r = 0.12 m vo = 260 m/s vf = 0 m/s = - 4506.7 N

Work Done by a Constant Force Friction acts on moving object m m f d If the net work is positive, the kinetic energy increases. If the net work is negative, the kinetic energy decreases.

You are pulling 100kg crate by a rope at 40° for 15m across a surface with friction. What is the net Work done on the box if it was initially sliding at 1m/s and ended up with a speed of 3m/s? 400 J 200 J 150 J 10 J Can’t be determined

You are pulling 100kg crate that was initially moving at 1m/s by a rope at 40° for 15m across a surface with friction, causing it to now have a speed of 3m/s. Is it possible to determine the Work done by Tension on the box with the given information? Yes No

You are pulling 100kg crate by a rope at 40° for 15m You are pulling 100kg crate by a rope at 40° for 15m. What would cause the net Work done on the box to be negative if it was initially sliding at 1m/s? fk > FTx If the crate slowed down Both of A & B Net Work can’t be negative b/c it doesn’t have direction

A child starts at rest at the top of a slide and slides down going pretty fast at the bottom. What are the signs of the Work done by individual forces, the Net Work, and Change in Kinetic Energy during that process? Wfk WFg WFN Wnet ΔKE A. + + + + + B. – + 0 + + C. – – 0 + + D. – + 0 – + – – – – + None of these

Work done by gravity Work and Energy m mg h m Object falls in a gravitational field Work and Energy m Work done by force (F) = mg h Kinetic energy = m Work done by gravity = Work done against friction = Wg = mgh

Can you solve #13 by: Work energy theorem Conservation of energy kinematics