Transcript Part III

Sect. 7.5: Kinetic Energy
Work-Kinetic Energy Theorem
• 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
Object undergoes displacement
Δr = Δx i (Δx= xf - xi) & velocity
change (Δv= vf -vi) under action
xi
xf
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
Combine (4) & (3):

a = [(vf)2 - (vi)2]/(2Δx)
Wnet = (½)m[(vf)2 - (vi)2]
(4)
(5)
• 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!
• 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.
• 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
Examples
vi = 20 m/s
vf = 30 m/s
m = 1000 kg
Conceptual
vi = 60 km/h
vf = 0
Δx = 20 m
vi = 120 km/h
Δx = ??
vf = 0
Example 7.6
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
Conceptual Example 7.7