WORK
In order for work to be done, three things are necessary:
•There must be an applied force.
•The force must act through a certain distance, called the
displacement.
•The force must have a component along the displacement.
Work is a scalar quantity
equal to the product of
the magnitudes of the
displacement and the
component of the force in
the direction of the
displacement.
W=F.r
W = F cos  r
UNITS: N.m, this unit is
called a Joule (J)
If the force acting on an object varies in magnitude and/or
direction during the object’s displacement, graphical analysis
can be used to determine the work done. F is plotted on the yaxis and the distance through which the object moves is plotted
on the x-axis. The work done is represented by the area under
the curve.
6.1 What work is done by a 60 N force in dragging the block a distance of
50 m when the force is transmitted by a rope making an angle of 30 with
the horizontal?
F = 60 N, 30˚
r = 50 m
W = F cos  r
= 60 cos 30˚ (50)
= 2598 J
6.2 How much work is done against gravity in lifting a 3.0 kg object
through a vertical distance of 40 cm.
m = 3 kg
h = 0.4 m
Force needed is equal to Fg
W = mgh
= 3(9.8)(0.4)
= 12 J
6.3 A push of 200 N moves a 100 N block up a 30inclined plane. The
coefficient of kinetic friction is 0.25 and the length of the plane is 12 m.
a. Find the work done by each force acting on the block.
Fa = 200 N
Fg = 100 N
θ = 30˚
μ = 0.25
r = 12 m
Forces acting: Ff Fa Fg and FN
FN does NO work.
Ff = μ FN
= μ Fgcos 30˚
= 0.25 (100) cos 30˚ = 21.6 N
W Fa = Fa r
= 200 (12)
= 2400 J
WFf = -Ff r
= - 21.6 (12)
= -259.2 J
FN
Fgy
θ
Ff
Fg
Fg
x
WFg = Fg r = -Fgxr = - Fgsin30˚r
= - 100 sin 30˚ (12)
= - 600 J
b. Show that the net work done by these forces is the same as the work of
the resultant force.
Net work:
ΣW = 2400 - 259.2 - 600
= 1540.8 J
FN
Fgy
θ
Ff
Fg
Fg
x
The resultant force:
ΣFx = Fa - Ff - Fgx
= 200 - 21.6 - 50
= 128.4 N
WF = F R . r
= 128.4 (12)
= 1540.8 J
ENERGY
Energy is that which can be converted into work. When
something has energy, it is able to perform work or, in a
general sense, to change some aspect of the physical world.
In mechanics we are concerned with two kinds of energy:
KINETIC ENERGY: K.E, energy possessed by a body by
virtue of its motion.
K
1
mv 2
Units: Joules (J)
2
POTENTIAL ENERGY: PE, energy possessed by a system by
virtue of position or condition.
PE = m g h
Units: Joules (J)
WORK-ENERGY PRINCIPLE:
The work of a resultant external force on a body is equal to the
change in kinetic energy of the body.
W =  KE
Units: Joules (J)
6.4 Find the kinetic energy of a 3200 N automobile traveling at 75 km/h?
Fg = 3200 N
v = 20.8 m/s
m = Fg/g = 326.5 kg
KE = ½ mv2
= ½ (326.5) (20.8)2
= 7.06 x104 J
6.5 What average force F is necessary to stop a 16 g bullet traveling at
260 m/s as it penetrates into wood at a distance of 12 cm?
m = 0.016 kg
x = 0.12 m
vo = 260 m/s
vf = 0 m/s
W = ΔKE
1
m(v 2f  vo2 )
2
1
Fr   mvo2
2
mvo2
(0.016)( 260) 2
F 

= - 4506.7 N
2(0.12)
2r
6.6 A 250 g object is held 200 mm above a workbench that is 1 m above the
floor. Find the potential energy relative to
a. the bench top
m = 0.25 kg
h = 0.2 m
b. the floor
h = 1.2 m
PE = mgh
= 0.25 (9.8) (0.2)
= 0.49 J
PE = mgh
= 0.25 (9.8) (1.2)
= 2.94 J
W = PE
CONSERVATIVE AND NON-CONSERVATIVE FORCES
The work done by a conservative force depends only on the
initial and final position of the object acted upon. An example
of a conservative force is gravity. The work done equals the
change in potential energy and depends only on the initial and
final positions above the ground and NOT on the path taken.
Friction is a non-conservative force and the work done in
moving an object against a non-conservative force depends on
the path. For example, the work done in sliding a box of books
against friction from one end of a room to the other depends on
the path taken.
LAW OF CONSERVATION OF ENERGY
The law of conservation of energy states that:
"Energy is neither created nor destroyed."
Energy can be transformed from one kind to another,
but the total amount remains constant.
For mechanical systems involving
conservative forces, the total
mechanical energy equals the
sum of the kinetic and potential
energies of the objects that make
up the system.
6.7 A 40 kg ball is pulled to one side until it is 1.6 m above its lowest point.
What will its velocity be as it passes through its lowest point?
m = 40 kg
h = 1.6 m
PE = K
mgh = ½ mv2
v  2 gh  (2)(9.8)(1.6) = 5.6 m/s
6.8 In an Atwood machine the two masses are 800 g and 700 g. The system
is released from rest. How fast is the 800-g mass moving after it has fallen
120 cm?
m1 = 0.8 kg
m2 = 0.7 kg
h = 1.2 m
ΔPE = ΔK
ΔPE = m2gh – m1gh
= 0.7 ( 9.8) (1.2) - 0.8 (9.8) (1.2)
= - 1.17 J
loss of PE = gain in K
1
K  mT (v 2f  v o2 ) = 1.17 J
2
2(1.17)
2K

vf 
= 1.25 m/s
0.7  0.8
m1  m2
6.9 If friction forces are negligible and the bead has a speed of 200 cm/s at
point A, a. What will be its speed at point B?
A
C
vA = 2 m/s
hA = 0.8 m
hB = 0 m
B
At point A
Energy: PEA + KA
At point B
PEA + KA = KB
1
1
2
mgh  mv A  mv B2
2
2
v B  v A2  2 gh
 (2) 2  2(9.8)(0.8)
= 4.4 m/s
b. What will be its speed at point C?
hC = 0.5 m
A
C
At point C
PEA + KA = PEC + KC
B
1
1
2
mghA  mv A  mghC  mvC2
2
2
1 2
vC  2( g (hA  hC )  v A )  2(9.8(0.8  0.5)  1 (2) 2 ) = 3.14 m/s
2
2
GENERAL CASE
In real life applications, some of the mechanical energy is lost
due to friction. The work due to non-conservative forces is
given by:
WNC = Δ K + Δ PE
6.10 Suppose the bead in Prob. 6.9 has a mass of 15 g and a speed of 2 m/s
at A, and it stops as it reaches point C. The length of the wire from A to C is
250 cm. How large is the average friction force that opposes the motion of
the bead?
m = 0.015 kg
vA = 2 m/s
hA = 0.8 m
r = 2.5 m
vC = 0 m/s
hC = 0.5 m
WFf = Δ K + Δ PE = Ff r
PE  mg (hC  hA ) = 0.015(9.8)(0.5-0.8) = - 0.04 J
1
K  m(vC2  v A2 ) = 0 - ½ (0.015)(2)2 = - 0.03 J
2
PE  K ( 0.04)  ( 0.03)
Ff 

= - 0.028 N
r
2 .5
6.11 A 64 N block rests initially at the top of a 30 m plane inclined at an
angle of 30. If k = 0.1. Find the final velocity at the bottom of the plane
from energy considerations.
Fg = 64 N
m = 64/9.8 = 6.5 kg
h = 30 m
μ = 0.1
r = 15 m
r
h
At the TOP
PE = mgh
= 64(15)
= 960 J
r = 30 sin 30˚= 15 m
Fg = 64 N
m = 64/9.8 = 6.5 kg
h = 30 m
μ = 0.1 r = 15 m
At the BOTTOM
PE = K + WFf
Ff = μ FN = μ Fg cos 30˚
= 0.1 (64 cos 30˚ )
= 5.54 N
K = PE - WFf
= 960 - 166.2
= 793.8 J
WFf = Ff r
= 5.54 (15)
= 166.2 J
K = ½ mv2
2 K  2(793.8)
= 15.6 m/s
v
6.5
m
POWER
Is the rate at which work is performed.
P=W =Fr = Fv
t
t
P = work/time
Units: J/s: watt (W)
6.12 An advertisement claims that a certain 1200-kg car can accelerate
from rest to a speed of 25 m/s in a time of 8.0 s. What power must the
motor produce to cause this acceleration?
m = 1200 kg
vo = 0 m/s vf = 25 m/s
t=8s
1
W = ΔK  m(v 2f  vo2 ) = ½ (1200)(25)2= 3.75x105 J
2
5
W 3.75 x10
= 47 kW

P
t
8
6.13 A 0.25 hp motor is used to lift a load at the rate of 5 cm/s. How great a
load can it lift at this constant speed? (1 hp = 746 W)
P = 0.25 hp (746 W/hp) = 186.5 W
v = 0.05 m/s
P = Fv
P 186.5
F 
= 3730 N
v
0.05
F 3730
m 
= 380.6 kg
9 .8
g
ELASTIC POTENTIAL ENERGY
Elastic potential energy is associated with elastic materials. The
force Fp applied to a spring to stretch it or to compress it an
amount x is directly proportional to x. That is: Fp = k x
Where k is a constant called the spring constant and is a
measure of the stiffness of the particular spring. The spring
itself exerts a force in the opposite direction:
Fs = - k x
Units: Newtons (N)
This force is sometimes called restoring force because the spring
exerts its force in the direction opposite to the displacement.
This equation is known as the spring equation or Hooke’s Law.
The elastic potential energy is given by:
PEs = ½ kx2 Units: Joules (J)
6.14 A dart of mass 0.100 kg is pressed against the spring of a toy dart gun.
The spring (k = 250 N/m) is compressed 6.0 cm and released. If the dart
detaches from the spring when the spring reaches its normal length, what
speed does the dart acquire?
m = 0.1 kg
k = 250 N/m
x = 0.06 m
PEs = K
½ kx2 = ½ mv2
v
250(0.06) 2
kx 2

= 3 m/s
0.1
m
6.15 A 0.20 kg ball is attached to a vertical spring as in the figure. The
spring constant k is 28 N/m. The ball, supported initially so that the spring
is neither stretched nor compressed, is released from rest. How far does the
ball fall before being brought to a momentary stop by the spring?
m = 0.20 kg
k = 28 N/m
vo = 0 m/s
ho = 0.5 m
PEg = PEs
mgho = ½ kx2
2mgho
2(0.2)(9.8)(0.5)
x
= 0.26 m

k
28