Lesson 1 on kinetic energy

1. Lesson 3
Kinetic Energy & the
Work-Energy Relation
Nelson Reference Pages:Nelson Reference Pages:
230- 232230- 232

2.  If a force is applied to a non-moving hockey puck (If a force is applied to a non-moving hockey puck (on aon a
horizontal surfacehorizontal surface), the puck will move in the direction of), the puck will move in the direction of
the applied force.the applied force.
 For the brief period of time that the force is applied, thereFor the brief period of time that the force is applied, there
will be displacement in the same direction as the force.will be displacement in the same direction as the force.
This means that positive work has been done.
 Work represents a transfer of energy so the puck mustWork represents a transfer of energy so the puck must
have gained energy, more specifically it gainedhave gained energy, more specifically it gained kinetic
energy (EK) because it is now moving.because it is now moving.
 Our understanding of EOur understanding of EKK came from experiments done bycame from experiments done by
Christian Huygens,, a mathematician and physicist, abouta mathematician and physicist, about
350 years ago. He found that the sum of the values of350 years ago. He found that the sum of the values of
mvmv22
, for moving billiard balls, before and after a collision, for moving billiard balls, before and after a collision
were equal. This quantity was changed to 1/2 mvwere equal. This quantity was changed to 1/2 mv22
andand
became known as kinetic energy. So,became known as kinetic energy. So,
EK = 1/2 mv2
, m is mass in kg,is mass in kg, v is speed in m/s, inis speed in m/s, in EK inin
JoulesJoules

3. Kinetic Energy and the Work- EnergyKinetic Energy and the Work- Energy
TheoremTheorem
 W = FW = FParallelParallel ΔdΔd
 But, F = maBut, F = ma
 So, W = ma ΔdSo, W = ma Δd
 But, a = (vBut, a = (v22 – v– v11)/(Δt))/(Δt)
 So, W = m [(vSo, W = m [(v22 – v– v11)/(Δt)] Δd)/(Δt)] Δd
 But, Δd = [(vBut, Δd = [(v22 + v+ v11)/2] Δt)/2] Δt
 So, W = m [(vSo, W = m [(v22 – v– v11)/(Δt)] [((v)/(Δt)] [((v22 + v+ v11)/2) Δt])/2) Δt]
 Or simplified,Or simplified, W = ½ mv2
2
- ½ mv1
2
 And,And, W = Δ EK
 This tells us that the change inThis tells us that the change in EK equals theequals the
work done.work done.

4.  Although we used kinematics equations, itAlthough we used kinematics equations, it
is important to remember that energy isis important to remember that energy is
not a vector quantity.not a vector quantity. However, work can
be either positive or negative.
 Example
 A 1.0 kg puck, initially at rest, is hit by a
hockey stick which applies a 12.0 N force
over a linear distance of 0.10 m. If the puck
slides on a frictionless horizontal surface
determine the final speed of the hockey puck.
Ans. 1.5 m/s

5. Practice Questions
Nelson TB Problems:
Page 235 #1, 2Page 235 #1, 2 (part b use one method only)(part b use one method only)
Page 263 #26, 29, 31, 63 (a, b)Page 263 #26, 29, 31, 63 (a, b)

6. Practice Questions
Nelson TB Problems:
Page 235 #1, 2Page 235 #1, 2 (part b use one method only)(part b use one method only)
Page 263 #26, 29, 31, 63 (a, b)Page 263 #26, 29, 31, 63 (a, b)