Work
Contents:
•Definition
•Tricky Bits
•Whiteboards
Work
- Transfer of energy
Work = (Force)(Distance)
W = Fd cos
F

Fcos
d
W = Fd cos
F

Fcos
d
Tricky bits :
0o cos = 1
180o cos = -1
(examples of zero, positive, negative work)
Whiteboards:
Work
1|2|3
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) = 139.9 Nm
= 139.9 Joules = 140 J
So a Nm = J (Joule)
140 J
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
84.6 N
Myron Wondergaim 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
2.5 m
Work
- Transfer of energy
Work = (Force)(Distance)
W = Fd
F = mg (if you are lifting)
F = μmg (if you are dragging)
W = Fd
F = mg (lifting) OR F = μmg (dragging)
Art Zenkraft does 342 J of work lifting a
12.0 kg mass how far? (2.91 m)
(J) W
(N) F
(m) d
(kg) m
( ) μ
W = Fd
F = mg (lifting) OR F = μmg (dragging)
Shirley Nott does 552 J of work sliding a
45.0 kg mass a distance of 10.4 m. What
must be the coefficient of friction? (0.120)
(J) W
(N) F
(m) d
(kg) m
( ) μ
Whiteboards:
Work and Weight and Friction
1|2|3
Helena Handbasket lifts a 5.2 kg box
from the floor to a 1.45 m tall shelf.
What work does she do?
F = mg
F = (5.2 kg)(9.8 N/kg) = 50.96 N
= 73.892 J
74 J
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 = 2012.712 N
F = mg
2012.712 N = m(9.8 N/kg), m = 205.4 kg
205.4 kg
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) = 51.45 N
W = Fd cos = Fd
137 J = (51.45 N)d, d = 2.66 m
2.66 m
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) = 183.75 N
W = Fd = (183.75 N)(34 m) = 6247.5 J ≈ 6250 J
6250 J
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) = 65.856 N
W = Fd
1200 J = (65.856 N)d, d ≈ 18.2 m
18.2 m
Work and Weight
Lifting things
W = Fd cos
sometimes
F = mg (weight)
F = mg = (4.0 kg)(9.8 N/kg) = 39.2 N
W = Fd = (39.2 N)(1.5 m) = 58.8 J
m= 4.0 kg
d = 1.5 m
Work and Friction
Dragging things
W = Fd cos
sometimes
F = μmg
F = μmg = (.26)(5.0 kg)(9.8 N/kg) = 12.74 N
W = Fd = (12.74 N)(3.5 m) = 44.59 J
d = 3.5 m
m= 5.0 kg
μ = .26