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 energy of a system increases
and a – sign for work when the energy of the system
decreases.
 Calculation of work done by various forces
 For Fig. 3a 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. 3b above:
 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 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 (Θ
+ 90^{o}) = mgs sin Θ
 Work done by f = fs cos f,s
= fs cos 180^{o} = fs
 Practice problems in 104
Problem Set for Work and Energy: 26, 8, 9.
 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.
 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.
 For two point masses, m_{1} and m_{2}
separated by a distance r, the potential energy of the
system is U = Gm_{1}m_{2}/r. The zero
point of the gravitational potential energy for two point
particles is infinity.
 You will find later that a mass at the end of a spring
of constant k, the elastic potential energy function U
= 1/2 kx^{2}, where x is the displacement from
the equilibrium position. The equilibrium position,
x = 0, is taken as the zero of potential
energy.
 Workenergy theorem. The work done by the net
force equals the change in the kinetic energy
 Example. Each graph in Fig. 4 below describes the onedimensional
motion of a 5.00 kg system during a 4.00 s interval. For
each case, what is the work done on the system by the
net force during the interval?
 In Fig. 4a, 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. 4b, 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(5.0 kg)(0)^{2} 1/2(5.0 kg)(5
m/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(5.0 kg)(5 m/s)^{2} 1/2(5.0 kg)(0)^{2
} = +62.5 J
 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 a nonconservative force acts on the system, work
done by the nonconservative force = (U_{f} +
K_{f})  (U_{i} + K_{i})
 Example: 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
 Repeat when µ_{k} between the object
and the plane is 1/4
 Solution:
 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}
or v_{f} = (60)^{1/2} m/s = 7.7 m/s
 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 = 1 kg v_{f}’^{2} 
60 J
40 J = 40 N  m = 40 kg m^{2}/s^{2}
= v_{f}’^{2 }kg
v_{f}’ = (40)^{1/2} m/s = 6.3
m/s
 Practice problems in 104
Problem Set for Work and Energy: 1023.
 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^{o}
= 0 = k • j
 Practice problem in 104
Problem Set for Work and Energy: 17.

