1. Sect. 7.5: Kinetic Energy Work-Kinetic Energy Theorem

2. Energy  The ability to do work Kinetic Energy  The energy of motion
“Kinetic”  Greek word for motion
An object in motion has the ability to do work

3. Wnet = ∑FΔx (2). (N’s 2nd Law in energy form!)
Object undergoes displacement
Δr = Δx i (Δx= xf - xi) & velocity
change (Δv= vf -vi) under action
of const. net force ∑F figure 
Text derivation
Calculus not needed! Instead,
Newton’s 2nd Law ∑F = ma (1). Work by const. force
W = FΔx (F,Δx in same direction).  Net (total) work:
Wnet = ∑FΔx (2). (N’s 2nd Law in energy form!)
Or using N’s 2nd Law: Wnet = maΔx (3). ∑F is constant
Acceleration a is constant  Ch. 2 kinematic equation:
(vf)2 = (vi)2 + 2aΔx  a = [(vf)2 - (vi)2]/(2Δx) (4)
Combine (4) & (3): Wnet = (½)m[(vf)2 - (vi)2] (5) xi xf
Figure 7.12: An object undergoing a displacement Δr = Δx î and a change in velocity under the action of a constant net force ∑F.

4. Wnet = (½)m(vf)2 - (½)m(vi)2  K (I) WORK-KINETIC ENERGY THEOREM
Summary: Net work done by a constant net force in accelerating an object of mass m from vi to vf is:
Wnet = (½)m(vf)2 - (½)m(vi)2  K (I)
DEFINITION: Kinetic Energy (K). (Kinetic = “motion”)
K  (½)mv2 (units are Joules, J)
WORK-KINETIC ENERGY THEOREM
Wnet = K = Kf - Ki ( = “change in”)
NOTE: The Work-KE Theorem (I) is 100% equivalent to N’s 2nd Law. IT IS Newton’s 2nd Law in work & energy language! We’ve shown this for a 1d constant net force. However, it is valid in general!

6. Work-Kinetic Energy Theorem
Net work on an object = Change in KE.
Wnet = K  (½)[m(vf)2 - m(vi)2]
Work-Kinetic Energy Theorem
Note: Wnet = work done by the net (total) force.
Wnet is a scalar.
Wnet can be positive or negative
(because KE can be both + & -)
Units are Joules for both work & K.

7. Moving hammer can do work on nail.
For hammer:
Wh = Kh = -Fd
= 0 – (½)mh(vh)2
For nail:
Wn = Kn = Fd
= (½)mn(vn)2 - 0

8. Examples Conceptual vi = 20 m/s vf = 30 m/s m = 1000 kg vi = 60 km/h
Δx = 20 m
vi = 120 km/h
vf = 0
Δx = ??

10. Example 7.6 Work-Kinetic Energy Theorem
A block, mass m = 6 kg, is pulled
from rest (vi = 0) to the right by a
constant horizontal force F = 12 N.
After it has been pulled for Δx = 3 m,
find it’s final speed vf.
Work-Kinetic Energy Theorem
Wnet = K  (½)[m(vf)2 - m(vi)2] (1)
If F = 12 N is the only horizontal force,
we have Wnet = FΔx (2)
Combine (1) & (2):
FΔx = (½)[m(vf)2 - 0]
Solve for vf: (vf)2 = [2Δx/m]
(vf) = [2Δx/m]½ = 3.5 m/s
Figure 7.13: (Example 7.6) A block pulled to the right on a frictionless surface by a constant horizontal force.

11. Conceptual Example 7.7
Figure 7.14: (Conceptual Example 7.7) A refrigerator attached to a frictionless, wheeled hand truck is moved up a ramp at constant speed.