1. Work
Contents:
Definition
Tricky Bits
Whiteboards

2. Work - Transfer of energy
Work = (Force)(Distance)
W = Fd cos F d 
Fcos
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3. W = Fd cos Fcos  Tricky bits : 0o cos = 1 180o cos = -1 F d
(examples of zero, positive, negative work)
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4. Whiteboards:
Work
1 | 2 | 3
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5. Fred O’Dadark exerts 13.2 N on a rope that makes a 32o angle with the ground, sliding a sled 12.5 m along the ground. What work did he do?
W = Fd cos = (13.2 N)(12.5 m) cos(32o) = Nm
= Joules = 140 J
So a Nm = J (Joule) W
140 J

6. Jane Linkfence does 132 J of work lifting a box 1. 56 m
Jane Linkfence does 132 J of work lifting a box 1.56 m. What is the weight of the box?
W = Fd cos
 = 0o (assuming she lifts straight up)
W = Fd, F = W/d = (132 Nm)/(1.56 m) = 84.6 N W
84.6 N

7. Bobbie Sachs exerts 43 N of force, and does 108 J of work pushing a sofa how far? (2 hints)
W = Fd cos
 = 0o (assume)
W = Fd, d = W/F = (108 Nm)/(43 N) = 2.5 m W
2.5 m

8. Work and Weight Lifting things
W = Fd cos
sometimes
F = mg (weight)
m= 4.0 kg
d = 1.5 m
F = mg = (4.0 kg)(9.8 N/kg) = 39.2 N
W = Fd = (39.2 N)(1.5 m) = 58.8 J
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9. Work and Friction Dragging things
W = Fd cos
sometimes
F = μmg
F = μmg = (.26)(5.0 kg)(9.8 N/kg) = N
W = Fd = (12.74 N)(3.5 m) = J
μ = .26
d = 3.5 m
m= 5.0 kg
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10. Whiteboards:
Work and Weight
1 | 2 | 3
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11. Helena Handbasket brings a 5. 2 kg box down from a 1. 45 m tall shelf
Helena Handbasket brings a 5.2 kg box down from a 1.45 m tall shelf. What work does she do? (3 hints)
F = mg
W = Fd cos
 = 180o (tricky tricky)
F = (5.2 kg)(9.8 N/kg) = N
W = Fd cos(180o) = (50.96 N)(1.45 m) cos(180o)
= J = -74 J
(The box does work on her. gives her energy) W
-74 J

12. Paul E. Wannacracker does 2375 J of work lifting what mass a height of 1.18 m?
F = mg
W = Fd cos = Fd
2375 J = F(1.18 m), F = N
N = m(9.8 N/kg), m = kg W
205.4 kg

13. Tubi O’ Notubi does 137 J of work lifting a 5
Tubi O’ Notubi does 137 J of work lifting a 5.25 kg mass to what height?
F = mg
F = (5.25 kg)(9.8 N/kg) = N
W = Fd cos = Fd
137 J = (51.45 N)d, d = 2.66 m W
2.66 m

14. Hugh Jass drags a 125 kg sled with a coefficient of kinetic friction of .15 a distance of 34 m. What work does he do?
F = μmg
F = (.15)(125)(9.8) = N
W = Fd = ( N)(34 m) = J ≈ 6250 J W
6250 J

15. Seymour Butz does 1200 J of work dragging a 32 kg box with a coefficient of kinetic friction of .21 how far?
F = μmg
F = (.21)(32)(9.8) = N
W = Fd
1200 J = ( N)d, d ≈ 18.2 m W
18.2 m