1. BELLWORK 2/24/2015 What does the work-energy theorem state?
Kinetic energy is proportional to the mass of the moving object. If it moved at the same speed, an arrow twice as massive as this one would have twice as much kinetic energy

2. ANSWER
The net work done on an object is equal to the change in its kinetic energy.

3. Work-Kinetic Energy Theorem
Net work done on a particle equals the change in its kinetic energy (KE)
Wnet = ΔKE
Wnet = KEf - KEi
F x d (cos θ) = ½ m(v)2f
Kinetic Books 7.7
Consider the foot kicking the soccer ball in Concept 1. We want to relate the work done by the force exerted by the foot on the ball to the ball’s change in kinetic energy. To focus solely on the work done by the foot, we ignore other forces acting on the ball, such as friction.
Initially, the ball is stationary. It has zero kinetic energy because it has zero speed. The foot applies a force to the ball as it moves through a short displacement. This force accelerates the ball. The ball now has a speed greater than zero, which means it has kinetic energy. The work-kinetic energy theorem states that the work done by the foot on the ball equals the change in the ball’s kinetic energy. In this example, the work is positive (the force is in the direction of the displacement) so the work increases the kinetic energy of the ball.
As shown in Concept 2, a goalie catches a ball kicked directly at her. The goalie’s hands apply a force to the ball, slowing it. The force on the ball is opposite the ball’s displacement, which means the work is negative. The negative work done on the ball slows and then stops it, reducing its kinetic energy to zero. Again, the work equals the change in energy; in this case, negative work on the ball decreases its energy.

4. EXAMPLE
A student is pushed along a frictionless horizontal surface by a friend with a constant force of 55N. How far must the student be pushed, starting from rest, so that her final Kinetic Energy is 450J?
Given:
Fpush = 55N
Vi = 0m/s; therefore
KEi = 0J
KEf = 450J

5. WORK Given: Fpush = 55N Vi = 0m/s; therefore KEi = 0J KEf = 450J Find:
Fornula:
Wnet = KEf – KEi
Wnet = F x d (cos0)
The force and the displacement are in the same direction, so theta =0
Wnet = 55N x d

6. WORK Fornula: Wnet = KEf – KEi Wnet = F x d (cos0)
The force and the displacement are in the same direction, so theta =0
Wnet = 55N x d
Wnet = KEf – Kei
55N x d = 450J – 0J
Now, solve for d
55N (d) = 450J
d = 450J
55N
d = 8.2 m

7. CLASSWORK
Complete Pg. 168 #’s 1-2, and 4