1. Chapter 5
Work and Energy

2. 5-1: Work
Work: the product of the magnitude of the component of force along the direction of displacement and the displacement

3. Work = parallel force x displacement
Equation: W = F// x d = F d cos
Symbol for work = W , SI Unit = joule
1J = 1kgm2/s2

4. Work is a scalar
(+) if the force & displacement are in the same direction
(-) if they are in opposite directions

5. Work done: Why? Energy change? Simple Examples:+
F & d same direction
increase
Pull or push something (even at an angle) … increases KE
Raise or lift something … increases PE 
F & d opposite direction
decrease
Friction … decreases KE
Lower something … decreases PE
F & d are perpendicular, or no net force
none
overall/total
You carry something … energy stays constant
Something falls … energy is conserved: PE → KE
Something is thrown up … energy is cons: KE → PE

6. Identify the work done in each case:1 2 3 4 5

7. 5-2: Energy We define energy as the ability to do work.
Work is the energy transferred to or from a system by a force that acts on it.
Video: Rube Goldberg Machine…

8. Energy Symbol: E Scalar Units: J, Joule cal, calorie
kcal, kilocalorie (Cal)

9. There are different forms of energy

10. KE = ½m(v) 2, same unit, joule
Kinetic Energy: energy of an object due to its motion
KE depends on the objects mass and speed, it’s given by:
KE = ½m(v) 2, same unit, joule

11. Problem 1
A bird of mass 2kg is traveling at a speed of 6m/s. What is its kinetic energy? (A: 36J) If the speed of the bird doubles what happens to its kinetic energy?
36 J
Image Credit: National Geographic

13. Problem 2
A 5 kg object has 250 J of KE. What is the speed of the object?
KE = ½mv2
2 KE = mv2
v2 = 2 KE /m
v = √(2 (250 J/ 5 kg)) = 10 m/s

14. Question:
Can kinetic energy be negative?

15. Potential Energy: energy associated with an object due to its position relative to a reference point
Symbol: U or PE
Units: Joule
There are different forms of potential energy

16. Gravitational potential energy: energy associated with an object due to its position relative to the Earth
PEg = mgh m = mass g = acceleration due to gravity
h = height

17. Sample Problem
A flower pot of mass 5kg is located 15m above the ground. A)What is its potential energy with respect to the ground? B) What work was done to raise it to its position?

18. Elastic potential energy: the potential energy in a stretched or compressed elastic object
PE = ½kx2
k = spring constant  x = displacement from equilibrium

19. 5-3: Conservation of Energy
Mechanical energy is the sum of kinetic energy and all forms of potential energy
ME = Σ KE + ΣPE
Conservation of Energy: energy can neither be created nor destroyed. It can only change from form to form.
PEi + KEi = PEf + KEf

20. Conservation of Energy

21. Ex: A skier with a mass of 88 kg hits a ramp of snow at 16 m/s
Ex: A skier with a mass of 88 kg hits a ramp of snow at 16 m/s. What is his total energy at this point? Et = KE = ½mv2 Et = ½ (88 kg) (16m/s)2 = 11, 264 J
Assume he becomes airborne and reaches a height of 3.7 m above the ground. He is now moving at 13.5 m/s. What is his total energy at this point?
Et = KE + PE
= ½mv2 + mgh
= ½ (88 kg) (13.5 m/s)2 +(88 kg)(9.8 m/s2)(3.7m)
= J J
= 11, 210 J

23. What happens when friction is present?
When friction is present, the work done by the frictional force W=fd is transferred away as heat energy.
DE = Wnc, the change in energy is equal to the work done by the non-conservative force.

24. A Force is “Non-Conservative” if:
“the work it does on an object that moves between two points depends on the path taken.”
“the work it does on an object that moves through a round trip is non-zero.”
Example: friction, tension, normal force, propulsion forces.

25. A Force is “Conservative” if:
“ the work it does on an object that moves between two points depends only on the position of these two points and not on the path.”
“the work it does on an object that moves through a round trip is zero.”
Example: gravity, force of a spring.

26. Ex: A clerk pulls a 300 kg package a distance of 10
Ex: A clerk pulls a 300 kg package a distance of 10.0 m at a constant speed. The coefficient of friction between the crate and the floor is 0.5. How much work does he do?

27. Given: m = 300 kg, d = 10 m, m=0.5
W = Fclerk d cosq N f
Fclerk W

28. ΣFx = 0 Σ Fy = 0 Fclerk – f = 0 N – W = 0 Fclerk = f N = W = mg f = m N Fclerk = m N = m mg = 0.5 (300 kg) (9.8m/s2) = 1470 N W = Fclerk d cosq = 1470 N ( 10m) cos 0º = 14,700 J

29. 5 – 4: Work, energy and power
Work- kinetic energy theorem states that the net work done on an object is equal to the change in kinetic energy of the object Wnet =  KE
Consider:

30. Since the force is related to the acceleration of the object by F=ma.
During the displacement, the force does work given by
Work = Fd
Since the force is related to the acceleration of the object by F=ma.
Work = Fd = (ma)d
Now the distance an object moves while accelerating from rest is
d = vit + ½at2
d = ½ at2

31. So the work can be written as
W = ma (½ at2) = ½m(at) 2
From the equations of motion
vf= vi + at (vi = 0)
So the work is
W = ½m(vf) 2 ; ie…the change in Kinetic energy!

32. Ex: A 10. 0 kg shopping cart is pushed from rest by a 250
Ex: A 10.0 kg shopping cart is pushed from rest by a N force against a 50.0 N friction force for a distance of 10.0 m. Find the speed of the cart.

33. For the friction force: W = (50.0N)(10.0m)cos180º= -500J
W = Fdcos
For the applied force:
W = (250.0N)(10.0m)cos 0º = 2,500J
For the friction force:
W = (50.0N)(10.0m)cos180º= -500J
Wt = 2,500 J + (-500J) = 2,000J
Wt = KE = KEf – KEi
KEf = ½mv2
v = √(2 KE /m)
= √ (2 ( 2,000J/ 10.0kg))
= 20.0 m/s

34. Power: the rate at which energy is transferred
Power = work/ time
P = W t
SI unit = watt, W
1 watt = 1 joule/second
Another common unit = horsepower, hp
1 hp = 746 W

35. Alternatively, since W = Fd
P = Fd/t
But d/  t = v so we can write the power as:
P = Fv

36. Ex: Two horses pull a cart Each exerts a force of 250
Ex: Two horses pull a cart Each exerts a force of N at a speed of 2.0 m/s for 10.0 min. Calculate the power provided by the horses.
t = 10.0 min x 60 s/min = 600 s
P = Fv = (250.0 N)(2.0 m/s) = 500 J/s, for each horse
Pt = 1000 W
How much work is done by the horses?
P = W/ t
W = P t
= (1000 W)(600s) = 6.0 x 105 J