Unit 4: Conservation of Energy. This is the halfway point! Chapters 1- 6 covered kinematics and dynamics of both linear and rotational motion. Chapters 7-12 focus on the conservation laws of energy, momentum, and angular momentum and their applications.
q = 0 and cosq=1
Example pushing the couch with 500N for 2m. The work would be 1000 Joules
Holding a 20kg mass object:
The force you apply equals mg=196N
W=Fdcosq = F(0)=0
Walking forward at constant velocity:
Force = 196N
d is nonzero
But cosq = cos90 = 0
W=Fdcosq = Fd(0)=0Limiting Cases
A 50-kg crate is pulled 40m by a 100N-force acting at a 37o angle. The force of friction is 50N
Determine the work done by the pulling, frictional, and net forces.Example 1: Work on a Crate
The free body diagram shows 4 forces:
And the work:
Note that the force pulling the mass does positive work and the force of friction does negative work.
The net work is he sum or +1200J
FG = mg = 147N
The force of the hiker holding the pack aloft FH = mg = +147N
The work done by the hiker is:
WH= FH(d)(cosq) = FHh= 147N(10.0m)= 1470J
Only the height matters -not the distance traveled.
The work done by gravity is:
WG= FG(d)(cos(180-q))= FG(d)(-cosq)=-FG(d)(cosq)=-FGh= -147N*(10.0m)=-1470J
Once again only the height matters -not the distance traveled.
The net work is just the sum of the work done by both forces or 0.
The sum can be graphically represented if a varying force.
Each shaded rectangle is one element of the sum.
The curve represents the function Fcosq.
Then, the work approximates the area under the curve.
As we make Dli smaller and smaller, the sum of rectangles gives a better and better estimate of the area under the curve.
In fact as it approaches zero we get an exact result for the area and for the work: