Download Kinetic Energy and Work

Chapter 7
Kinetic Energy and Work
Introduction to Energy
 The concept of energy is one of the
most important topics in science
 Every physical process that occurs in
the Universe involves energy and
energy transfers or transformations
Work and Energy
Energy: scalar quantity associated with a state (or
condition) of one or more objects.
Work and energy are scalars, measured in N·m or
Joules, J, 1J=kg∙m2/s2
Energy can exist in many forms - mechanical,
electrical, nuclear, thermal, chemical….
Energy Approach to Problems
 The energy approach to describing
motion is particularly useful when the
force is not constant
 An approach will involve Conservation
of Energy
• This could be extended to biological
organisms, technological systems and
engineering situations
Work and Energy
Energy is a conserved quantity - the total
amount of energy in the universe is constant.
Energy can be converted from one type to
another but never destroyed.
Work and energy concepts can simplify solutions of
mechanical problems - they can be used in an
alternative analysis
Systems
 A system is a small portion of the
Universe
• We will ignore the details of the rest of the
Universe
 A critical skill is to identify the system
Valid System
 A valid system may
•
•
•
•
be a single object or particle
be a collection of objects or particles
be a region of space
vary in size and shape
Environment
 There is a system boundary around the
system
• The boundary is an imaginary surface
• It does not necessarily correspond to a
physical boundary
 The boundary divides the system from
the environment
• The environment is the rest of the Universe
Work
 The work, W, done on a system by an
agent exerting a constant force on the
system is the product of the magnitude, F,
of the force, the magnitude Δr of the
displacement of the point of application of
the force, and cos θ, where θ is the angle
between the force and the displacement
vectors
Work

W = F Δr cos θ
• The displacement is that of the point of
application of the force
• A force does no work on the object if the
force does not move through a displacement
• The work done by a force on a moving object
is zero when the force applied is
perpendicular to the displacement of its point
of application
Work Example
 The normal force, n,
and the gravitational
force, mg, do no work
on the object
cos θ = cos 90° = 0
 The force F does do
work on the object
More About Work
 The system and the environment must be

determined when dealing with work
 The environment does work on the system
(Work by the environment on the system)
The sign of the work depends on the direction of
F relative to Δr
 Work is positive when projection of F onto Δr
is in the same direction as the displacement
 Work is negative when the projection is in the
opposite direction
Work Is An Energy Transfer
• This is important for a system approach
to solving a problem
• If the work is done on a system and it is
positive, energy is transferred to the
system
• If the work done on the system is
negative, energy is transferred from the
system
Work done by a constant force
Work done on an object by a constant force is
defined to be the product of the magnitude of
the displacement and the component of the
force parallel to the displacement
W = FII · d
Where FII is the component of the force F
parallel to the displacement d
Work
In other words W = F d cosq
Where θ is the angle between F and d
F
q
d
If θ is > 90o, work is negative. A decelerating car has
negative work done on it by its engine.
The unit of work is called Joule (J), 1 J = 1 N·m
1J=kg∙m2/s2
Scalar Product of Two Vectors
 The scalar product

of two vectors is
written as A . B
It is also called
the dot product
A . B = A B cos θ
θ is the angle
between A and B
Scalar Product
 The scalar product is commutative
A .B = B .A
 The scalar product obeys the distributive law
of multiplication
A . (B + C) = A . B + A . C
Dot Products of Unit Vectors

î  î  ĵ  ĵ  k̂  k̂  1
î  ĵ  î  k̂  ĵ  k̂  0
 Using component form with A and B:
A  A x î  A y ĵ  A zk̂
B  B x î  B y ĵ  B zk̂
A  B  A xB x  A yB y  A zB z
Work - on and by
A person pushes block 30 m along the
ground by exerting force of 25 N on the
trolley. How much work does the person do
on the trolley?
W = F d = 25N x 30m = 750 Nm
Trolley does -750 Nm work on the person
Ftp
Fpt
d


A force F  4x iˆ  3y ˆj N acts on an object as
the object moves in the x direction from the origin
to x = 5.00 m. Find the work W = Fdr done on
the object by the force.


A force F  4x iˆ  3y ˆj N acts on an object as
the object moves in the x direction from the origin
to x = 5.00 m. Find the work W = Fdr done on
the object by the force.
f
W   F  dr 
i
5m

0
5m

0


ˆ 3yˆ
ˆ
4xi
j N  dxi
2 5m
x
 4 N m  xdx 0   4 N m  2
0
 50.0 J
Mechanical Energy
Mechanical energy (energy associated with
masses) can be thought of as having two
components: kinetic and potential
– Kinetic energy is energy of motion
– Potential energy is energy of position
Kinetic Energy
In many situations, energy can be
considered as “the ability to do work”
Energy can exist in different forms
Kinetic energy is the energy associated
with the motion of an object
A moving object can do work on another
object
– E.g hammer does work on nail.
Kinetic Energy
Consider an object with mass m moving in a
straight line with initial velocity vi. To accelerate it
uniformly to a speed vf a constant net force F is
exerted on it parallel to motion over a distance d.
v 2f  v i2  2a(x f  xi )  v i2  2ad
a
v 2f  v i2
2d
Work done on object
W=Fd=mad
So
W m
v 2f  v i2
2d
(NII)
d
Kinetic Energy
If we rearrange this we obtain
1 2 1 2
W  mv f  mv i
2
2
We define the quantity ½mv2 to be the
translational kinetic energy (KE) of the object
W = DKE
This is the ‘Work-Energy Theorem’:
“The net work done on an object is equal
to its change in kinetic energy”
The Work-Energy Theorem
W = DKE
• Note
– The net work done on an object is the work
done by the net force.
– Units of energy and work must be the
same (J)
Energy: scalar quantity associated with a state (or
condition) of one or more objects.
I. Kinetic energy.
II. Work.
III. Work - Kinetic energy theorem.
IV. Work done by a constant force
- Gravitational force
V. Work done by a variable force.
- Spring force.
1D-Analysis
3D-Analysis
VI.
Power
I. Kinetic energy
Energy associated with the state of motion of an object.
1 2
K  mv
2
Units:
1 Joule = 1J = 1 kgm2/s2
II. Work
Energy transferred “to” or “from” an
object by means of a force acting on the
object.
To  +W
From  -W
- Constant force:
v 2f
 v02
Fx  max
 2a x d  a x 
v 2f  v02
2d
1
2
2 1
 Fx  max  m(v f  v0 )
2
d
1
 m(v 2f  v02 )  K f  Ki  Fx d  W  Fx d
2
Work done by the force = Energy transfer due to the force.
-To calculate the work done on an object by a force during
a displacement, we use only the force component along
the object’s displacement. The force component perpendicular
to the displacement does zero work.
 
W  Fx d  Fd cos   F  d
Assumptions:
1) F = constant force
2) Object is particle-like (rigid object, all
parts of the object must move together).
A force does +W when it has a vector component in the
same direction as the displacement, and –W when it has
a vector component in the opposite direction. W=0 when it
has no such vector component.
  90  W
180    90  W
  90  0
Net work done by several forces = Sum of works done by
individual forces.
Calculation: 1) Wnet= W1+W2+W3+…
2) Fnet  Wnet=Fnet d
II. Work-Kinetic Energy Theorem
DK  K f  K i  W
Change in the kinetic energy of the
particle = Net work done on the particle
III. Work done by a constant force
- Gravitational force:
W  mgd cos 
Rising object: W= mgd cos180º = -mgd 
Fg transfers mgd energy from the object’s kinetic energy.
Falling down object: W= mgd cos 0º = +mgd 
Fg transfers mgd energy to the object’s kinetic energy.
External applied force + Gravitational force:
DK  K f  K i  Wa  Wg
Object stationary before and
after the lift: Wa+Wg=0
The applied force transfers the same amount of
energy to the object as the gravitational force
transfers from the object.
IV. Work done by a variable force
- Spring force:


F  kx
Hooke’s law
k = spring constant  measures spring’s stiffness. Units: N/m
Hooke’s Law
 When x is positive


(spring is stretched), F is
negative
When x is 0 (at the
equilibrium position), F is
0
When x is negative
(spring is compressed),
F is positive
Hooke’s Law
 The force exerted by the spring is
always directed opposite to the
displacement from equilibrium
 F is called the restoring force
 If the block is released it will oscillate
back and forth between –x and x
Hooke’s law:
1D  Fx  kx
xi
Δx
x
Work done by a spring force:
•
•
Assumptions: •
•
Spring is massless  mspring << mblock
Ideal spring  obeys Hooke’s law exactly.
Contact between the block and floor is frictionless.
Block is particle-like.
- Calculation:
1) The block displacement must be divided
into many segments of infinitesimal width, Δx.
2) F(x) ≈ cte within each short Δx segment.
Fx
xi
Δx
xf
x
Fj
Ws   F j Dx  Dx  0 

xf
xi
xf
xf
xi
xi
F dx   (kx) dx  k 
xf
xi
 1  2 2
   k ( x f  xi )
 2 
Ws>0  Block ends up closer to the
relaxed position (x=0) than it was initially.
Ws<0  Block ends up further away from
x=0.
if xi  0
Ws=0  Block ends up at x=0.
1
1
2
Ws  k xi  k x 2f
2
2
1
Ws   k x 2f
2
 
 1  2
x dx    k  x
 2 
Work done by an applied force + spring force:
DK  K f  K i  Wa  Ws
Block stationary before and after the displacement:
Wa= -Ws  The work done by the applied force displacing
the block is the negative of the work done by the spring force.
If it takes 4.00 J of work to stretch a Hooke'slaw spring 10.0 cm from its unstressed
length, determine the extra work required to
stretch it an additional 10.0 cm.
If it takes 4.00 J of work to stretch a Hooke'slaw spring 10.0 cm from its unstressed
length, determine the extra work required to
stretch it an additional 10.0 cm.
1
2
4.00 J k 0.100 m   k  800 N m
2
1
2
DW   800 0.200  4.00 J 12.0 J
2
A 2.00-kg block is attached to a spring of force
constant 500 N/m. The block is pulled 5.00 cm to the
right of equilibrium and released from rest. Find the
speed of the block as it passes through equilibrium if
(a) the horizontal surface is frictionless and (b) the
coefficient of friction between block and surface is
0.350.
A 2.00-kg block is attached to a spring of force
constant 500 N/m. The block is pulled 5.00 cm to the
right of equilibrium and released from rest. Find the
speed of the block as it passes through equilibrium if
(a) the horizontal surface is frictionless and (b) the
coefficient of friction between block and surface is
0.350.


1 2 1 2 1
2 2
W s  kxi  kxf   500 5.00  10
 0  0.625 J
2
2
2
1 2 1 2 1 2
W s  m vf  m vi  m vf  0
2
2
2
2  W 
2 0.625
vf 
m

2.00
m s  0.791 m s
A 2.00-kg block is attached to a spring of force
constant 500 N/m. The block is pulled 5.00 cm to the
right of equilibrium and released from rest. Find the
speed of the block as it passes through equilibrium if
(a) the horizontal surface is frictionless and (b) the
coefficient of friction between block and surface is
0.350.
1 2
1 2
m vi  fkDx  W s  m vf
2
2
1 2
0   0.350 2.00 9.80  0.050 0 J 0.625 J m vf
2
1
0.282 J  2.00 kg v2f
2
vf 
2 0.282
m s  0.531 m s
2.00
Work done by a general variable force:
• Assume that during a
very small displacement,
Δx, F is constant
• For that displacement,
W ~ F Δx
• For all of the intervals,
xf
W   Fx Dx
xi
Work done by a general variable force:
xf
•
lim
Dx 0
 F Dx  
x
xf
xi
xi
Fx dx
• Therefore,
xf
W   Fx dx
xi
• The work done is
equal to the area
under the curve
3D-Analysis

F  Fxiˆ  Fy ˆj  Fz kˆ ;

dr  dx iˆ  dy ˆj  dz kˆ
Fx  F ( x), Fy  F ( y ), Fz  F ( z )
rf
xf
yf
zf
 
dW  F  dr  Fx dx  Fy dy  Fz dz  W   dW   Fx dx   Fy dy   Fz dz
ri
xi
Work-Kinetic
Energy Theorem - Variable force
x
x
f
f
xi
xi
W   F ( x)dx   ma dx
dv
dv
ma dx  m
dx  m v dx  mvdv
dt
dx
 dv dv dx dv 
 v
 
 dt dx dt dx 
vf
vf
vi
vi
W   mv dv  m  v dv 
1 2 1 2
mv f  mvi  K f  K i  DK
2
2
yi
zi
V. Power
Time rate at which the applied force does work.
- Average power: amount of work done
in an amount of time Δt by a force.
-Instantaneous power:
-instantaneous time rate of doing work.
Pavg
W

Dt
dW
P
dt
Units: 1 watt= 1 W = 1J/s
1 kilowatt-hour = 1000W·h = 1000J/s x 3600s = 3.6 x 106 J
= 3.6 MJ
 
dW F cos  dx
 dx 
P

 F cos     Fv cos   F  v
dt
dt
 dt 
F
φ
x
In the figure below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m
across a frictionless floor. Find an expression for the speed vf at
the end of that distance if the block’s initial velocity is: (a) 0 and
(b) 1m/s to the right.
(a)
N
Fy
Fx
mg
N
Fx
Fy
mg
In the figure below a 2N force is applied to a 4kg block at a
downward angle θ as the block moves right-ward through 1m
across a frictionless floor. Find an expression for the speed vf at
the end of that distance if the block’s initial velocity is: (a) 0 and
(b) 1m/s to the right.
(a)
N
Fy
Fx
Fx
Fy
mg
mg
(a) v0  0  DK  0.5mv2f
(2 N )(1m) cos q  0.5(4kg)v
 v f  cos q m / s
 
W  F  d  ( F cos q )d
N
W  DK  0.5m(v 2f  v02 )
(b) v0  1m / s  DK  0.5mv2f  2 J
2
f
(2 N ) cos q  0.5(4kg)v 2f  2 J
 v f  1  cos q m / s
(c) The situation in fig.(b) is similar in that the block is initially
moving at 1m/s to the right, but now the 2N force is directed
downward to the left. Find an expression for the speed of the
block at the end of the 1m distance.
N
Fy
Fx
mg
N
Fx
Fy
mg
(c) The situation in fig.(b) is similar in that the block is initially
moving at 1m/s to the right, but now the 2N force is directed
downward to the left. Find an expression for the speed of the
block at the end of the 1m distance.
N
Fy
Fx
N
Fx
Fy
mg
mg
 
W  F  d  ( F cos q )d
W  DK  0.5m(v 2f  v02 )
(b) v0  1m / s  DK  0.5mv 2f  2 J
(c) v0  1m / s  DK  0.5mv2f  2 J
(2 N )(1m) cos q  0.5(4kg)v 2f  2 J
 (2 N ) cos q  0.5(4kg)v 2f  2 J
 v f  1  cos q m / s
 v f  1  cos q m / s
A small particle of mass m is pulled to the top of a
frictionless half-cylinder (of radius R) by a cord that passes
over the top of the cylinder, as illustrated in Figure. (a) If the
particle moves at a constant speed, show that F = mgcos.
(Note: If the particle moves at constant speed, the
component of its acceleration tangent to the cylinder must
be zero at all times.) (b) By directly integrating W = Fdr,
find the work done in moving the particle at constant speed
from the bottom to the top of the half-cylinder.
A small particle of mass m is pulled to the top of a frictionless
half-cylinder (of radius R) by a cord that passes over the top of
the cylinder, as illustrated in Figure. (a) If the particle moves at
a constant speed, show that F = mgcos. (Note: If the particle
moves at constant speed, the component of its acceleration
tangent to the cylinder must be zero at all times.) (b) By directly
integrating W = Fdr, find the work done in moving the particle
at constant speed from the bottom to the top of the halfcylinder.
(a)
 Fx  m ax
F  m g cosq  0
F  m g cosq
A small particle of mass m is pulled to the top of a frictionless
half-cylinder (of radius R) by a cord that passes over the top of
the cylinder, as illustrated in Figure. (a) If the particle moves at
a constant speed, show that F = mgcos. (Note: If the particle
moves at constant speed, the component of its acceleration
tangent to the cylinder must be zero at all times.) (b) By directly
integrating W = Fdr, find the work done in moving the particle
at constant speed from the bottom to the top of the halfcylinder.
f
(b)
W   F  dr
 2
W 

0
i
 2
m g cosq Rdq  m gR sin q 0
W  m gR  1 0  m gR
(We use radian measure to express the next bit of displacement
dr as in terms of the next bit of angle moved through dθ: dr=Rdθ)
Two springs with negligible masses, one with spring
constant k1 and the other with spring constant k2, are
attached to the endstops of a level air track as in Figure. A
glider attached to both springs is located between them.
When the glider is in equilibrium, spring 1 is stretched by
extension xi1 to the right of its unstretched length and
spring 2 is stretched by xi2 to the left. Now a horizontal
force Fapp is applied to the glider to move it a distance xa to
the right from its equilibrium position. Show that in this
process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b)
the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is
related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by
the force Fapp is (k1 + k2)xa2.
Two springs with negligible masses, one with spring
constant k1 and the other with spring constant k2, are
attached to the endstops of a level air track as in Figure. A
glider attached to both springs is located between them.
When the glider is in equilibrium, spring 1 is stretched by
extension xi1 to the right of its unstretched length and
spring 2 is stretched by xi2 to the left. Now a horizontal
force Fapp is applied to the glider to move it a distance xa to
the right from its equilibrium position. Show that in this
process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b)
the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is
related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by
the force Fapp is (k1 + k2)xa2.
f
(a)
W 1   F1dx 
i
(b)
W2
 xi2  xa

 xi2
xi1  xa

xi1
k1x dx 




1 
1
2
k1  xi1  xa   xi21   k1 xa2  2xaxi1
 2
2 
1 
1
2
2 
k2x dx  k2   xi2  xa   xi2  k2 xa2  2xaxi2
 2
2 
Two springs with negligible masses, one with spring
constant k1 and the other with spring constant k2, are
attached to the endstops of a level air track as in Figure. A
glider attached to both springs is located between them.
When the glider is in equilibrium, spring 1 is stretched by
extension xi1 to the right of its unstretched length and
spring 2 is stretched by xi2 to the left. Now a horizontal
force Fapp is applied to the glider to move it a distance xa to
the right from its equilibrium position. Show that in this
process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b)
the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is
related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by
the force Fapp is (k1 + k2)xa2.
(c) Before the horizontal force is applied, the springs exert equal
forces:
kx
k1xi1  k2xi2
xi2 
1 i1
k2
Two springs with negligible masses, one with spring
constant k1 and the other with spring constant k2, are
attached to the endstops of a level air track as in Figure. A
glider attached to both springs is located between them.
When the glider is in equilibrium, spring 1 is stretched by
extension xi1 to the right of its unstretched length and
spring 2 is stretched by xi2 to the left. Now a horizontal
force Fapp is applied to the glider to move it a distance xa to
the right from its equilibrium position. Show that in this
process (a) the work done on spring 1 is k1(xa2+2xaxi1) , (b)
the work done on spring 2 is k2(xa2 – 2xaxi2) , (c) xi2 is
related to xi1 by xi2 = k1xi1/k2, and (d) the total work done by
the force Fapp is (k1 + k2)xa2.
(d)
1
1
2
W 1  W 2  k1xa  k1xaxi1  k2xa2  k2xaxi2
2
2
k1xi1
1
1
2
2
 k1xa  k2xa  k1xaxi1  k2xa
2
2
k2
1
  k1  k2  xa2
2
N6. A 2kg lunchbox is sent sliding over a frictionless surface,
in the positive direction of an x axis along the surface.
Beginning at t=0, a steady wind pushes on the lunchbox in the
negative direction of x. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the
force from the wind do on the lunch box from t=1s to t=5s?
N6. A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface.
Beginning at t=0, a steady wind pushes on the lunchbox in
the negative direction of x. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the
force from the wind do on the lunch box from t=1s to t=5s?
Motion  concave downward
parabola
xt
1 2
t
10
dx
2
 1 t
dt
10
dv
2
a
   0.2m / s 2
dt
10
v
F  cte  ma  (2kg)( 0.2m / s 2 )  0.4 N
(a) t  1s  v f  0.8m / s
W  F  x  (0.4 N )(t  0.1t 2 )
K f  0.5(2kg)(0.8m / s ) 2  0.64 J
N6. A 2kg lunchbox is sent sliding over a frictionless surface, in
the positive direction of an x axis along the surface. Beginning
at t=0, a steady wind pushes on the lunchbox in the negative
direction of x, Fig. below. Estimate the kinetic energy of the
lunchbox at (a) t=1s, (b) t=5s. (c) How much work does the force
from the wind do on the lunch box from t=1s to t=5s?
Motion  concave downward
parabola
xt
1 2
t
10
dx
2
 1 t
dt
10
dv
2
a
   0.2m / s 2
dt
10
v
F  cte  ma  (2kg)( 0.2m / s 2 )  0.4 N
W  F  x  (0.4 N )(t  0.1t 2 )
(c) W  DK  K f (5s)  K f (1s)  0  0.64  0.64 J
(b) t  5s  v f  0
K f  0J
N12. In the figure below a horizontal force Fa of magnitude
20N is applied to a 3kg book, as the book slides a distance of
d=0.5m up a frictionless ramp. (a) During the displacement,
what is the net force done on the book by Fa, the gravitational
force on the book and the normal force on the book? (b) If the
book has zero kinetic energy at the start of the displacement,
what is the speed at the end of the displacement?
x
y
N
Fgx
mg
Fgy
N12. In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m
y
up a frictionless ramp. (a) During the
x
displacement, what is the net force done
N
on the book by Fa, the gravitational force
F
on the book and the normal force on the
F
mg
book? (b) If the book has zero kinetic
energy at the start of the displacement,
 
what is the speed at the end of the
N  d W  0
displacement?
Only Fgx , Fax do work
gx
gy
(a) W  WFa x  WFg x
or
Wnet
 
 Fnet  d
Fnet  Fax  Fgx  20 cos 30  mg sin 30
Wnet  (17.32 N  14.7 N )0.5m  1.31J
N12. In the figure below a horizontal force Fa of magnitude 20N
is applied to a 3kg book, as the book slides a distance of d=0.5m
y
up a frictionless ramp. (a) During the
x
displacement, what is the net force done
N
on the book by Fa, the gravitational force
F
on the book and the normal force on the
F
mg
book? (b) If the book has zero kinetic
energy at the start of the displacement,
 
what is the speed at the end of the
N  d W  0
displacement?
Only Fgx , Fax do work
gx
gy
(b) K 0  0  W  DK  K f
W  1.31J  0.5mv2f  v f  0.93m / s
N15. (a) Estimate the work done represented by the graph below
in displacing the particle from x=1 to x=3m. (b) The curve is given
by F=a/x2, with a=9Nm2. Calculate the work using integration
N15. (a) Estimate the work done represented by the graph below
in displacing the particle from x=1 to x=3m. (b) The curve is given
by F=a/x2, with a=9Nm2. Calculate the work using integration
(a) W  Area under curve 
(11.5squares)(0.5m)(1N )
 5.75 J
3
9
1
1
(b) W   2 dx  9   9(  1)  6 J
3
 x 1
1 x
3
N19. An elevator has a mass of 4500kg and can carry a maximum
load of 1800kg. If the cab is moving upward at full load at constant
speed 3.8m/s, what power is required of the force moving the cab
to maintain that speed?
Fa
mg
N19. An elevator has a mass of 4500kg and can carry a maximum
load of 1800kg. If the cab is moving upward at full load at constant
speed 3.8m/s, what power is required of the force moving the cab
to maintain that speed?
Fa
mtotal  4500kg  1800kg  6300kg




Fnet  Fa  mg  ma  0
 Fa  mg  (6300kg)(9.8m / s 2 )  61.74kN
mg
 
P  F  v  (61.74kN )(3.8m / s)  234.61kW
N17. A single force acts on a body that moves along an x-axis.
The figure below shows the velocity component versus time for
the body. For each of the intervals
v
AB, BC, CD, and DE, give the sign
B
C
(plus or minus) of the work done by
D
A
t the force, or state that the work is zero.
E
N17. A single force acts on a body that moves along an x-axis.
The figure below shows the velocity component versus time for
the body. For each of the intervals
v
AB, BC, CD, and DE, give the sign
B
C
(plus or minus) of the work done by
D
A
t the force, or state that the work is zero.
E

1
W  DK  K f  K 0  m v 2f  v02
2
AB  vB  v A  W  0
BC  vC  vB  W  0

CD  vD  vC  W  0
DE  vE  0, vD  0  W  0
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
(a) Pulley 1 : v  cte  Fnet  0  2T  mg  0  T  98 N
mg
Hand  cord : T  F  0  F 
 98 N
2
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
(b) To rise “m” 0.02m, two segments of the cord must be shorten
by that amount. Thus, the amount of the string pulled down at the
left end is: 0.04m
(c) WF  F  d  (98 N )(0.04m)  3.92 J
15E. In the figure below, a cord runs around two massless,
frictionless pulleys; a canister with mass m=20kg hangs from one
pulley; and you exert a force F on the free end of the cord.
(a) What must be the magnitude of F if
you are to lift the canister at a constant
P2
speed? (b) To lift the canister by 2cm,
how far must you pull the free end of
the cord? During that lift, what is the
T
T T
work done on the canister by (c) your
P1
force (via the cord) and (d) the
mg
gravitational force on the canister?
(d ) WFg  mgd  (0.02m)(20kg)(9.8m / s 2 )  3.92 J
WF+WFg=0
There is no change in kinetic energy.
Challenging problems – Chapter 7
Two trolleys of masses m1=400 kg and m2=200 kg are connected by a
rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes
to push them with a force of 1200N.
N1
From the point of view of kinetics,
N2
Situation 1
F1r F2r
does the relative position of the
F
trolleys matter? If the rod can
m1g
m2g
only stand an applied force of 500N,
which trolley should be up front?
Challenging problems – Chapter 7
Two trolleys of masses m1=400 kg and m2=200 kg are connected by a
rigid rod. The trolleys lie on a horizontal frictionless floor. A man wishes
to push them with a force of 1200N.
N1
From the point of view of kinetics,
N2
Situation 1
F1r F2r
does the relative position of the
F
trolleys matter? If the rod can
m1g
m2g
only stand an applied force of 500N,
which trolley should be up front?
F’2r
F’1r
Situation 1
Action and reaction forces: F1r force that the rod does on m1
F2r force that the rod does on m2
F’1r force that m1 does on rod.
F’2r force that m2 does on rod.
Bl . m1 : F  F1r  m1a  1200 N  F1r  400a
Bl . m2 : F2r  m2a  F2r  200a
a
F
 2m / s 2
(m1  m2 )
F1r  F2r  400 N
Rod negiglible mass, Fnet on rod=0
F’1r = F’2r  rigid rod does not deform
F1r = F’1r  Action / Reaction
Forces that the rod does on the
blocks as it tries to counteract the
deformation that F could induce.
Challenging problems – Chapter 7
Two trolleys of masses m1=400 kg and m2=200 kg are connected by a rigid
rod. The trolleys lie on a horizontal frictionless floor. A man wishes to push
them with a force of 1200N. From the point of view of kinetics, does the
relative position of the trolleys matter? If the rod can only stand an applied
force of 500N, which trolley should be up front?
F’2r
F’1r
Action and reaction forces:
Situation 2
Bl . m2 : F  F2r  m2a  1200 N  F2r  200a
Bl . m1 : F1r  m1a  F1r  F2r  400a
a
F
 2m / s 2
(m1  m2 )
F1r force that the rod does on m1
F2r force that the rod does on m2
F’1r force that m1 does on rod.
F’2r force that m2 does on rod.
Rod negiglible mass, Fnet on rod=0
F’1r = F’2r  rigid rod does not deform
F1r = F’1r  Action / Reaction
F1r  F2r  800 N
From
kinetic
point
of
view,
Situation1=Situation2same acceleration”.
Situation 2
From dynamics  F1r(1)≠F1r(2) Only situation (1) is
possible F1r=F’1r = 400N <Fmax=500N
N2
F2r
F
m2g
N1
F1r
m1g
2. A
car with a weight of 2500N working with a power of 130kW
develops a velocity of 112 km/hour when traveling along a
horizontal straight highway. Assuming that the frictional forces
(from the ground and air) acting on the car are constant but not
negligible: What is the value of the frictional forces?
(a) What is the car’s maximum velocity on a 50 incline hill? (b)
What is the power if the car is traveling on a 100 inclined hill at
36km/h?
2. A car with a weight of 2500N working with a power of 130kW develops a velocity of
112 km/hour when traveling along a horizontal straight highway. Assuming that the
frictional forces (from the ground and air) acting on the car are constant but not
negligible: What is the value of the frictional forces?
(a) What is the car’s maximum velocity on a 50 incline hill? (b) What is the power if the
car is traveling on a 100 inclined hill at 36km/h?
N
f
Situation 1
F
mg
P  130kW
v  112km / h  31.11m / s
 
P
130kW
P  F v  F  
 4178.7 N
v 31.11m / s
F  f  ma  0 (v  cte )  f  F  4178.7 N
2. A car with a weight of 2500N working with a power of 130kW develops a
velocity of 112 km/hour when traveling along a horizontal straight
highway. Assuming that the frictional forces (from the ground and air)
acting on the car are constant but not negligible: (a) What is the car’s
maximum velocity on a 50 incline hill? (b) What is the power if the car is
traveling on a 100 inclined hill at 36km/h?
Situation 2
F  mg sin q  f  0  F  (2500 N ) sin 5  4178.7 N  0
 F  4396.6 N
N
P 130000W
(a) v  
 29.56m / s
F 4396.6 N
f
Fgx
mg
(b) v  36km / h  10m / s, q  10
F  (2500 N ) sin 10  4178.7 N  0  F  4612.82 N
P  F  v  46128.2W
F
Fgy