 Phyllis Fleming Physics Physics 104
 Outline - Work and Energy
1. Definition of Work

1. 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.

2. Calculation of work done by various forces 1. For Fig. 3a above:

1. Work done by F = Fs cos f,s = Fs cos 0o = Fs
2. Work done by N = Ns cos N,s = Ns cos 90o = 0
3. Work done by mg = (mg)s cos mg,s = (mg)s cos -90o = 0
4. Work done by f = fs cos f,s = fs cos 180o = -fs

2. For Fig. 3b above:

1. Work done by F = Fs cos f,s = Fs cos Θ
2. Work done by N = Ns cos N,s = Ns cos 90o = 0
3. Work done by mg = (mg)s cos mg,s = (mg)s cos -90o = 0
4. Work done by f = fs cos f,s = fs cos 180o = -fs

3. For Fig. 3c above:

1. Work done by F = Fs cos f,s = Fs cos 0o = Fs
2. Work done by N = Ns cos N,s = Ns cos 90o = 0
3. Work done by mg = (mg)s cos mg,s
= (mg)s cos -(Θ + 90o) = -mgs sin Θ
4. Work done by f = fs cos f,s = fs cos 180o = -fs

3. Practice problems in 104 Problem Set for Work and Energy: 2-6, 8, 9.

2. Mechanical Energy

1. Kinetic Energy K is energy of motion. For an object of mass m moving with a velocity v, K = 1/2 mv2

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.

1. 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.

2. For two point masses, m1 and m2 separated by a distance r, the potential energy of the system is U = -Gm1m2/r. The zero point of the gravitational potential energy for two point particles is infinity.

3. You will find later that a mass at the end of a spring of constant k, the elastic potential energy function U = 1/2 kx2, where x is the displacement from the equilibrium position.  The equilibrium position, x = 0,  is taken as the zero of potential energy.

3. Work-energy theorem. The work done by the net force equals the change in the kinetic energy

1. Example. Each graph in Fig. 4 below describes the one-dimensional 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? 1. In Fig. 4a, the area under the a versus t graph, the change in velocity = 3.0 m/s2 (4.0 s) = 12 m/s. Work done by net force = change in kinetic energy = 1/2 mvf2 - 1/2 mvo2
= 1/2(5.0 kg)(12 m/s)2 - 1/2(5.0 kg)(0)2 = 360 J

2. In Fig. 4b, the velocity is constant. There is no change in kinetic energy and no work is done.

3. Work done by net force = change in kinetic energy
= 1/2 mvf2 - 1/2 mvo2
= 1/2(5.0 kg)(0)2 -1/2(5.0 kg)(5 m/s)2
= - 62.5 J

4. Work done by net force = change in kinetic energy
= 1/2 mvf2 - 1/2 mvo2
= 1/2(5.0 kg)(5 m/s)2 -1/2(5.0 kg)(0)2
= +62.5 J

4. The total energy of a system E = U + K

1. If no non-conservative force acts on a system, Ef = Ei
or (Uf + Kf) = (Ui + Ki)

2. If a non-conservative force acts on the system, work done by the non-conservative force = (Uf + Kf) - (Ui + Ki)

3. 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/s2,

1. use energy considerations to find the velocity of the object at the bottom of the incline

2. Repeat when µk between the object and the plane is 1/4

4. Solution:

1. 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 Uf = 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 vi is zero so that Ki = 1/2 mvi2 = 0. From conservation of energy, Ui + Ki= Uf + Kf
2.0 kg(10 m/s2)(3 m) + 0= 0 + 1/2(2.0 kg) vf2
60 m2/s2= vf2 or vf = (60)1/2 m/s = 7.7 m/s 2. At the bottom of the incline Uf = 0; at top of incline Ui = mgh.
Ki = 0. With the frictional force, a nonconservative force acting on the block, mechanical energy is not conserved.
(Fnet)y= may= m(0)
FN - mg cos Θ = 0
FN = mg cos Θ
fk = µkFN = µk mg cos Θ = µk mg(4/5)

Since the distance down the plane s = 5 m and cos fk,s = 180o, the work by friction = (fk) s cos fk,s
=
k mg cos Θ}s cos 180o
= (1/4)(2 kg)(10 m/s2)(4/5)}(5m)(-1) = -20 J.

Work by friction =   (Uf + Kf)    -    (Ui + Ki)
20 J = (0 + 1/2 mvf2) - (mgh + 0)
-20 J = 1/2(2 kg)vf2    - (2 kg)(10 m/s2)(3 m)
-20 J = 1 kg vf2           - 60 J
40 J = 40 N - m = 40 kg m2/s2 = vf2 kg
vf’ = (40)1/2 m/s = 6.3 m/s 5. Practice problems in 104 Problem Set for Work and Energy: 10-23.

3. Scalar or Dot Product 1. Definition C = A • B = |A||B|cos A, B

2. Work = Fs cos F,s may be written as F • s

3. Unit vectors

1. i • i = |i||i|cos i, i = |1||1|cos 0o = 1

j • j = |j||j|cos j, j = |1||1|cos 0o = 1

k • k = |k||k|cos k, k = |1||1|cos 0o = 1

2. i • j = |i||j|cos i, j = |1||1|cos 90o = 0 = j • i

i • k = |i||k|cos i, k = |1||1|cos 90o = 0 = k • i

j • k = |j||k|cos j, k = |1||1|cos 90o = 0 = k • j

4. Practice problem in 104 Problem Set for Work and Energy: 17.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming September 25, 2002 April 3, 2003