Outline  Work and Energy


 Definition of Work
 Work
= Fs cos F,s,
where F is the force, s is the distance moved and F,s
is the angle between F and s. Work is a scalar
quantity. Since work is a scalar quantity, you are free to
use a + sign for work when the work increases the energy of
a system and a  sign for work when the work takes energy
out of the system.
 Calculation of work done by various forces
(f is the frictional force)
 For Fig. 2a above
 Work done by F = Fs cos F,s
= Fs cos 0^{o} = Fs
 Work done by N = Ns cos N,s
= Ns cos 90^{o} = 0
 Work done by mg = (mg)s cos mg,s
= (mg)s cos 90^{o}= 0
 Work done by f = fs cos f,s
= fs cos 180^{o} = fs
 For Fig. 2b
 Work done by F = Fs cos F,s
= Fs cos Θ
 Work done by N = Ns cos N,s
= Ns cos 90^{o} = 0
 Work done by mg = (mg)s cos mg,s
= (mg)s cos 90^{o}= 0
 Work done by f = fs cos f,s
= fs cos 180^{o} = fs
 For Fig. 3c
 Work done by F = Fs cos F,s
= Fs cos 0^{o} = Fs
 Work done by N = Ns cos N,s
= Ns cos 90^{o} = 0
 Work done by mg = (mg)s cos mg,s
= (mg)s cos (Θ
+ 90^{o}) = mgs sin Θ
 Work done by f = fs cos f,s
= fs cos 180^{o}= fs
 Sample problems in 107
Problem Set for Work and Energy: 16.
 Workenergy theorem
The work done by the net force equals the change in the
kinetic energy. The kinetic energy K = 1/2 mv^{2}, where
m is the mass of the object and v its speed.
Example:
Each graph in Fig.3 below describes the onedimensional motion
and finds the work done in each case:
 In Fig. 3a, the area under the a versus t graph, the change
in velocity
= 3.0 m/s^{2 }(4.0 s) = 12 m/s. Work done by net force
= change in kinetic energy = 1/2 mv_{f}^{2}
 1/2 mv_{o}^{2} = 1/2(5.0 kg)(12 m/s)^{2}
 1/2(5.0 kg)(0)^{2} = 360 J
 In Fig. 3b, the velocity is constant. There is no change
in kinetic energy and no work is done.
 Work done by net force = change in kinetic energy =
1/2 mv_{f}^{2}  1/2 mv_{o}^{2}
= 1/2(5.0kg)(0)^{2} 1/2(5.0kg)(5m/s)^{2}
= 62.5 J
 Work done by net force = change in kinetic energy =
1/2 mv_{f}^{2}  1/2 mv_{o}^{2
} = 1/2(5.0kg)(5m/s)^{2} 1/2(5.0kg)(0)^{2}
= +62.5 J
 Sample problems in 107
Problem Set for Work and Energy: 7, 914.
 Mechanical Energy
 Kinetic Energy K is energy of motion. For an object of mass
m moving with a velocity v,
K = 1/2 mv^{2}
 Potential Energy U is energy of position. Potential energy
at point P equals the negative of the work done by a conservative
force in going from a point of zero potential to point P.
Every conservative force has its own potential energy function.
 Near the earth's surface, where the weight of the object
is mg, the gravitational potential energy function U =
mgh, where m is the mass of the object, g is the acceleration
due to gravity and h is the vertical height above the
zero potential energy point. You usually take the zero
of potential energy at the lowest point, unless a specific
location is given for U = 0 in the statement of the problem.
 Elastic potential energy function U = 1/2 kx^{2},
where k is the spring constant and x is the displacement
from the equilibrium position. The zero of potential energy
for the spring is when the mass is at the equilibrium
position, x = 0.
 The potential energy for the gravitational force F_{g
}= (Gm_{1}m_{2}/r^{2}), where
the force is in the negative radial direction, is
U =  Gm_{1}m_{2}/r. The zero for this
U is when r = infinity.
Type of conservative force

Force

Potential Energy Function
U

U = 0 point

Gravitational force near
Earth's surface = a constant

mg

mgh = mgy

y= h = 0

Gravitational force between
two "point" objects = a variable

F_{g} = Gm_{1}m_{2}r/r^{3}
F = Gm_{1}m_{2}/r^{2}

Gm_{1}m_{2}/r

r = ∞

Elastic force = a variable

F = kx

1/2 kx^{2}

x = 0

 The total energy of a system E = U + K.
 If no nonconservative force acts on a system,
E_{f} = E_{i}, or (U_{f}
+ K_{f}) = (U_{i} + K_{i})
 If nonconservative forces act on the system, work done
by the nonconservative forces =
(U_{f}
+ K_{f})  (U_{i} + K_{i})
Example:
(a) An object of mass m = 2.0 kg is released
from rest at the top of a frictionless incline of height 3
m and length 5 m. Taking g = 10 m/s^{2}, use
energy considerations to find the velocity of the object at
the bottom of the incline.
(b) Repeat when µ_{k} between the object and
the plane is 1/4.
Solution:
(a) Since the incline is frictionless and
no other nonconservative force acts on the object, energy
is conserved. Take the initial point i at the top of the incline
and the final point f at the bottom of the incline. Let U_{f
} = 0. At the initial point the potential energy is mgh,
where h is the vertical height above the bottom of
the incline. The object being released from the top of the
incline means that its initial velocity v_{i } is
zero so that K_{i} = 1/2 mv_{i}^{2 }
= 0. From conservation of energy,
U_{i} 
+ 
K_{i} 
= 
U_{f} 
+ 
K_{f} 
_{
} _{ } 2.0 kg(10 m/s^{2})(3 m) 
+ 
0 
= 
0 
+ 
1/2(2.0 kg) v_{f}^{2} 

60 m^{2}/s^{2}
= v_{f}^{2} 
v_{f} = (60)^{1/2}m/s
= 7.7 m/s 
(b) At the bottom of the incline
U_{f} = 0; at top of incline U_{i } = mgh.
K_{i } = 0. With the frictional force, a nonconservative
force acting on the block, mechanical energy is not conserved.
(F_{net})_{y}= ma_{y}= m(0)
F_{N}  mg cos Θ
= 0
F_{N} = mg cos Θ
f_{k }= µ_{k}F_{N} = µ_{k
}mg cos Θ = µ_{k
}mg(4/5)
Since the distance down the plane s = 5 m and cos f_{k}, s
= 180^{o}, the work by friction = (f_{k})
s cos f_{k},
s
= {µ_{k} mg cos Θ}s
cos 180^{o
}=^{ } (1/4)(2 kg)(10 m/s^{2})(4/5)}(5m)(1)
= 20 J.
Work by friction = (U_{f} + K_{f})  (U_{i
} + K_{i})
20 J = (0 + 1/2 mv_{f}’^{2})  (mgh
+ 0)
20 J = 1/2(2 kg)v_{f}’^{2}  (2 kg)(10
m/s^{2})(3 m)
20 J = 1kg v_{f}’^{2}  60 J
60 J  20 J = 1kg v_{f}’^{2
} 40 J = 40 Nm = 40 kg m^{2}/s^{2} =
v_{f}’^{2 }kg; v_{f}’
= (40)^{1/2} m/s = 6.3 m/s.
 Sample problems in 107
Problem Set for Work and Energy: 1421, 2341.
 Scalar or Dot Product
 Definition C = A • B
= ABcos
A, B
 Work = Fs cos F,s
may be written as F • s
 Unit vectors
 i • i = iicos
i,
i = 11cos 0^{o}
= 1
j • j = jjcos j,
j = 11cos 0^{o}
= 1
k • k = kkcos k,
k = 11cos 0^{o}
= 1
 i • j = ijcos
i,
j = 11cos 90^{o}
= 0 = j • i
i • k = ikcos i,
k = 11cos 90^{o}
= 0 = k • i
j • k = jkcos j,
k = 11cos 90^{0}
= 0 = k • j
 Sample problems in 107
Problem Set for Work and Energy: 8, 22.

