Unit 5
Work and Energy
5.1
Work and kinetic energy
5.2
Work - energy theorem
5.3
Potential energy
5.4
Total energy
5.5
Energy diagram of a mass-spring system
5.6
A general study of the potential energy curve
5.1
Work and kinetic energy
(1) Work: unit in joule (J)
y
Work done by the force F
x
W = Fx d = ( F cosθ )d = F ⋅ d
Fx : the force along the displacement
Remark:
Positive work is done on an object when the point of application of the force moves in the
direction of the force.
Mathematical statement: Work done is the dot product of the force and the displacement.
Dot product:
A ⋅ B = AB cosθ = Ax B x + Ay B y + Az Bz
Example
A 75.0-kg person slides a distance of 5.00m on a straight water slide, dropping through a
vertical height of 2.50 m. How much work does gravity do on the person?
θ
θ
h=2.5
mg
d=5.00m
1
Answer
The gravity along the displacement has a magnitude mg cos θ.
By the definition of work done, the work done of the gravity on the person is given by
W = (mg cos θ ) d = mg (h / d ) d = mgh = (75.0)(9.8)(2.50) = 1837.5 J
Remarks:
(1)
(h, h)
When F or d is not a constant (see the left figure,
the movement of the particle is always varying).
y
W = ∑ Fi ⋅ ∆si
∆wi = Fi ⋅ ∆s i ,
(0, 0)
i
where ∆si is extremely small. The work done of
C
x
the gravity on the ball’s falling down is given by
0
0
W =
∫ F ⋅ ds =∫ − Fˆj ⋅ (dx iˆ + dy ˆj ) =−F ∫ dy =−mg ∫ dy =mgh
C
(2)
C
h
h
Power: work done per unit time
∆W dW F ⋅ ds
.
=
=
∆t → 0 ∆t
dt
dt
Power = lim
Since v =
ds
, we have the power equals to the dot product of force and velocity. e.g.
dt
Power = F ⋅ v .
If the force and the velocity are in the same direction, P = Fv.
(3)
Kinetic energy: K ≡ 12 mv 2 .
5.2
Work - energy theorem
If an object is moving with an acceleration a and a distance d is moved, then according to the
formula v 2f = vi2 + 2ad , we have
F
v 2f = vi2 + 2( )d .
m
vi
vf
The work done on the object W is given by
W = Fd =
1
1
m 2
(v f − vi2 ) = mv 2f − mvi2 = K f − K i
2
2
2
K f − Ki =
∆K .
Or, we can rewrite it as W =
2
The work-energy theorem states that the total work done on the object by the forces equals
the change in kinetic energy.
Remark: A proof by using integration.
Suppose a particle moves in a straight line (1-D)
xf
W = ∫ F ( x )dx
xi
 F = ma = m
xf
∴ W = ∫ m(
xi
dv
,
dt
v=
dx
dt
xf
vf
dv
dx
)dx = ∫ mdv
= ∫ mvdv
xi
vi
dt
dt
= 21 mv 2f − 21 mvi2 = K f − Ki
Example
A 47.2-kg block of ice slides down an incline 1.62 m long and 0.902 m high. A worker
pushes up on the ice parallel to the incline so that it slides down at constant speed. The
coefficient of kinetic friction between the ice and the incline is 0.110. Find (a) the force
exerted by the worker, (b) the work done by the worker on the block of ice, and (c) the work
done by gravity on the ice.
N
F
Answer
(a)
0.902
Angle of inclination = θ
1.62m
fk
0.902
sin θ =
1.62
=
θ 33.8°
mg θ
The normal reaction, =
N 47.2 × 9.8 × cos 33.8
=
° 384.2 N
Frictional force, fk = 0.110 × 384.2 =
42.3 N
The forces which acts on the ice:
Force exerted by the worker + frictional force = down plane force = mg sin θ
Force exerted by the worker = 47.2 × 9.8sin 33.8o − 42.3
= 215.02 N
3
Work done by the worker = − 215.02 × 1.62
(b)
= − 348.33 J
(c)
Work done by the gravity = 47.2 × 9.8 × 0.902 =
417.2 J
5.3
Potential energy
(a)
The gravitational potential energy
U = mgh
The difference of the potential energy: ∆U = Uf − Ui = −W,
where W is the work done by gravity if the block is released.
Reference point
In this case, W is positive.
(b)
The elastic potential energy
Hooke’s law states that the restoring force is proportional to the elongation of the spring from
its natural length, e.g. F = −kx
where F is the restoring force and k is called spring constant.
F
F’
If the elongation goes from zero to x, the work done by the applied force F’ is given by
x
x
x
x
0
0
0
0
W = ∫ F ' dx = ∫ (− F )dx = ∫ kxdx = 12 kx 2
= 12 kx 2 .
Suppose the spring is stretched a distance x1 initially. Then the work we have to stretch it to a
greater elongation x2 is
x2
x2
x1
x1
2
2
W = ∫ Fdx = ∫ kxdx = 12 kx 2 − 12 kx1 .
When a spring is compressed by an amount x, the work done of the applied force
x
x
0
0
= ∫ Fdx = ∫ kxdx =
U=
1 2
kx
2
1 2
kx
2
(F: applied force)
(i.e. the elastic potential energy stored in the spring)
4
5.4
Total energy
E = K + U,
where E: mechanical energy, K: kinetic energy, U: potential energy.
For an isolated system (no frictional force), E = const. or it is conserved.
K + U = const.
This is the referred to as the conservation of mechanical energy.
Example
A 263-g block is dropped with an initial speed such that it hits onto a vertical
light spring of force constant k = 2.52 N/cm. The block sticks to the spring,
and the spring compresses 11.8 cm before coming momentarily to rest.
While the spring is being compressed, how much work is done (a) by the
force of gravity and (b) by the spring? (c) What is the speed of the block just
before it hits the spring?
Answer
Method I: By the work-energy theorm
(a)
Let the compression of the spring be x. The work done by the block’s weight is
W1 =mgx =( 263 × 10−3 ) ( 9.8 ) (11.8 × 10−2 ) =0.304 J.
(b)
The work done by the spring is
2
1
1
W2 =
− kx 2 =
− ( 2.52 × 102 ) (11.8 × 10−2 ) =
−1.74 J.
2
2
(c)
The speed vi of the block just before it hits the spring is given by
1
∆K = 0 − mvi2 = W1 + W2 ,
2
which yields
vi
=
( −2 )(W1 + W2 )
=
m
( −2 )( 0.304 − 1.74 )
3.3 m s
=
263 × 10−3
Method II: By the conservation of mechanical energy
Let the gravitational potential energy of the block be zero when it is at its lowest point. We
have K .E. + G.P.E. + E.P.E. =
constant .
1
1 2
1
1
Therefore, 2 𝑚𝑣𝑖2 + 𝑚𝑚𝑥 + 0 = 0 + 0 + 2 𝑘𝑥 2 . This equation gives mvi2 + mgx =
kx , and
2
2
thus the result for vi.
5
Example (Challenging)
A particle of mass m slides down the smooth inclined face of a wedge of mass 2m, and
inclination α, which is free to move in a smooth horizontal table. Use equations of
momentum and energy to obtain an expression
m
for the velocity of the particle relative to the
wedge when the particle has moved a relative
distance s from rest down the inclined face of the
2m
α
wedge.
Answer
Let the velocity of the wedge be V and that of the particle relative to the wedge be u. By
conservation of horizontal momentum, since there is no horizontal force acting on the system,
2mV − m (u cos α− V) = 0.
∴
1
V = u cos α .
3
By conservation of energy,
=
α
mgs sin
1
1
(2m)V 2 + m {(u cos α − V ) 2 + (u sin α ) 2 }
2
2
1
2
1
∴ gs sin α = u 2 cos 2α + u 2 cos 2 α + u 2 sin 2 α ,
9
9
2
1
V
m
V
u
α
2m
 6 gs sin α  2
u=
 .
2
 2 + sin α 
6
5.5
Energy diagram of a mass-spring system
A spring is fixed at one end and the other end is attached to mass m which moves back and
forth with an amplitude A
•
Elastic potential energy
U ( x ) = 21 kx 2 .
F
If frictional force is neglected, we have the energy of the system
being conserved, e.g. E = constant.
When x=0, U = 0, K = E and K = 21 mv 2 ,
v is the greatest velocity in the whole process of oscillation.
At x = −A, E = max(U), K = 0 ⇒
v = 0.
dU
= The negative slope of the curve U(x) against x,
dx
•
F=−
•
At the origin,
dU
= 0 , it is the equilibrium
dx
position. The resultant force is zero.
•
When x > 0,
dU
> 0 , hence F < 0 represents
dx
that the restoring force is pointing to the left.
•
When x < 0,
dU
< 0 , hence F > 0 represents
dx
that the restoring force is pointing to the right.
•
When x = 0, it is called equilibrium position.
7
5.6
A general study of the potential energy curve
The following curve shows a profile of an arbitrary potential energy with the position.
Unstable
equilibrium
Region I
Region II
Stable equilibrium
position
a
b
c
d
Conservation of mechanical energy
E = K + U(x) = const.
K = 21 mv 2 ≥ 0
•
If E = E0, the particle will stay at point x0, otherwise K has to be negative.
•
When E = E1, the motion of the particle is confined between x1 ≤ x ≤ x2, x1 and x2 are
called turning points.
•
When E = E2, depending on the initial condition of x, the particle will either move in
region I (a ≤ x ≤ b) or region II (c ≤ x ≤ d).
•
When E = E3, the particle will move between x = x3 and ∞.
•
Unstable equilibrium position: Any maximum point in a potential energy curve is an
unstable equilibrium position.
•
Stable equilibrium position: Any minimum point in a potential energy curve is a
stable equilibrium position.
8
Example (Challenging)
A particle of mass m is attached to one end of a light elastic string whose other end is fixed to
a point O on an inclined plane (inclination angle = 45o with the horizontal). The natural
length of the string is l and its force constant is mg/l. The particle is held on the inclined plane
so that the string lies just unstretched along a line of greatest slope and then released from
rest.
O
l
m
45o
(a)
Suppose that the inclined plane is smooth, determine the lowest position that the
particle can reach. Determine also the equilibrium position.
Suppose that the inclined plane is rough enough and the frictional coefficient is µ.
(b)
Show that if the particle stops after its first descending a distance d along the
greatest slope, then the distance is given by
=
d
2l (1 − µ ) .
Answer
O
l
m
s
45o
(a)
Note that the particle is momentarily stopped at its lowest position. By the
conservation of mechanical energy, the loss in gravitational energy becomes the gain in the
elastic potential energy.
1 mg 2
( ) s = mg ( s sin 45o ) ,
2 l
where s is the extension of the string when the particle is at its lowest position.
9
1 s
1
The above equation gives ( ) =
. Hence, we obtain s = 2l . The lowest position is
2 l
2
located at
l + s = l + 2l = (1 + 2)l from O.
For equilibrium of the particle, the down plane force of the particle balances with the elastic
force, we can write
mg sin 45o = (
mg
) seq .
l
O
l
m seq
45o
Hence, we obtain seq =
l + seq =+
l
l
2
l
. That is, the equilibrium position is located at
2
1
=
(1 +
2
)l from O.
(b)
O
l
m
d
Rough surface
45o
This part differs from part (a) by the frictional force. In the view of energy, we have
1 mg 2
( )d + m (mg cos 45o ) d =
mg (d sin 45o ) .
2 l
After simplification, we have
1 d
1
1
,
µ=
( )+
2 l
2
2
which gives
d
=
2l (1 − µ ) .
10