Phyllis Fleming Physics Physics 107
 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 work increases the energy of a system and a - sign for work when the work takes energy out of the system.

2. Calculation of work done by various forces
(f is the frictional force)

1. For Fig. 2a 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. 2b

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

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. Sample problems in 107 Problem Set for Work and Energy: 1-6.

2. Work-energy theorem

The work done by the net force equals the change in the kinetic energy. The kinetic energy K = 1/2 mv2, where m is the mass of the object and v its speed.

Example: Each graph in Fig.3 below describes the one-dimensional motion and finds the work done in each case:

1. In Fig. 3a, 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. 3b, 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(5.0kg)(0)2 -1/2(5.0kg)(5m/s)2 = -62.5 J

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

5. Sample problems in 107 Problem Set for Work and Energy: 7, 9-14.

3.  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. Every conservative force has its own potential energy function.

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. 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.

2. Elastic potential energy function U = 1/2 kx2, 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.

3. The potential energy for the gravitational force Fg = (Gm1m2/r2), where the force is in the negative radial direction, is
U = - Gm1m2/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 Fg = -Gm1m2r/r3   F = Gm1m2/r2 -Gm1m2/r r = ∞ Elastic force = a variable F = -kx 1/2 kx2 x = 0

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

4. If no non-conservative force acts on a system,
Ef = Ei, or (Uf + Kf) = (Ui + Ki)
5. If non-conservative forces act on the system, work done by the non-conservative forces =
(Uf + Kf) - (Ui + Ki)

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/s2,  use energy considerations to find the velocity of the object at the bottom of the incline.

(b) Repeat when µk 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 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 vf = (60)1/2m/s = 7.7 m/s

(b) At the bottom of the incline Uf = 0; at top of incline Ui = mgh. Ki = 0. With the frictional force, a non-conservative 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 fks = 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 = 1kg vf2 - 60 J
60 J - 20 J = 1kg vf2
40 J = 40 N-m = 40 kg m2/s2 = vf2 kg; vf’ = (40)1/2 m/s = 6.3 m/s.

6. Sample problems in 107 Problem Set for Work and Energy: 14-21, 23-41.

4. 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 900 = 0 = k • j

4. Sample problems in 107 Problem Set for Work and Energy: 8, 22.

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 Website Designed By: Questions, Comments To: Date Created: Date Last Updated: Susan D. Kunk Phyllis J. Fleming October 8, 2002 April 16, 2003