The following areas are covered in this review:

Terms:                                                        (back to the top)

Be able to define and properly apply the following terms.

• work
• power
• Joule
• Watt
• energy
• potential energy
• gravitational potential energy
• kinetic energy
• conservation of energy
• elasticity

Objectives:                                                        (back to the top)

You should be able to:

• calculate work done by a force
• express work quantities in proper units
• distinguish between positive and negative work
• calculate power when work is done
• express power quantities in proper units
• define energy
• distinguish between potential and kinetic energy
• calculate the gravitational potential energy of an object
• calculate the kinetic energy of an object
• express energy quantities in proper units
• calculate the total energy of an object
• relate work to changes in total energy
• identify transfers of energy
• apply energy conservatio to interactions between objects
• explain where "lost" energy goes in interactions
• define elasticity
• distinguish between completely elastic and inelastic interactions
• solve collision problems using conservation of momentum and kinetic energy

Work and Power                  (notes that apply)             (solutions)

Concepts

• Work is the product of force and displacement.  W = FD    The unit of work is Nm = Joule
• The force used must be along the direction of motion.  If the applied force is not in the propoer direction, only its component along the line of motion may be used.
• If the force is in the direction of motion it does positive work.  If the force is in the opposite direction of the motion it does negative work.
• If the force is perpendicular to the line of motion, or if there is no motion, no work is done by the force.
• Power is the rate at which work is done, or energy changes.  P = W / T   The unit of power is the watt.  1 W = 1 J / 1 s.

1.  How much work do you do when you hold a bag of groceries?

2.  How much work is done when a horizontal force of 20.0 N moves an object 4.0 m horizontally?

3.  When you pull your sled with a rope which makes an angle of 30° to the horizontal, how
     much work do you do if you pull with 25.0 N of force and the sled moves 25 m?

4.  You push the couch across the living room floor with a horizontal force of 200 N.
     The force of friction is 150 N.  What is the work done by

a)  you?
b)  friction?
c)  the weight of the couch?
d)  the net force on the couch?

6.  When you start to run your legs develope 1000 w of power during the 1.5 s of your
     acceeleration in which you cover 8.0 m.  How much force do your legs push on the ground.
 

Potential and Kinetic Energy         (notes that apply)             (solutions)

Concepts

• Energy is the ability of an object to exert forces and cause changes.
• Mechanical energy is derived from an object's position and/or state of motion.
• Gravitational potential energy is the energy an object possesses due to its vertical position.
• Gravitational potential energy is calculated from, PE = mgh, where h is the vertical distance from a reference height (PE = 0)
• Kinetic energy is the energy an object possesses due to its state of motion.
• Kinetic energy is calculated from, KE = 1/2mV².
• The unit of measurement for all energies is the Joule when standard units are used.
• The total energy of an object is the sum of its potential and kinetic energies.

1.  What is the gravitational potential energy of a 2.0 Kg book 1.5 m above the floor?

2.  If the gravitational potential energy of a 40.0 Kg rock is 500 J relative to the ground, how
     high is the rock?

3.  If the rock of #2 was lifted to twice its original height, how would the value of its potential
     energy change?

4.  What is the kinetic energy of a 1500 Kg car moving at 15 m/s?

5.  If the car in #4 changed its speed to 30 m/s, how would the value of its kinetic energy change?

6.  A 200 Kg tiger has a kinetic energy of 14400 J.  What is the speed of the tiger?

7.  A 1.5 Kg soccer ball has a speed of 20 m/s when it is 15 m above the ground.
    What is the total energy of the ball?
 

Changes in Energy               (notes that apply)      (solutions)

Concepts

• Work causes a change in the total energy of an object.
• Positive work increases the total energy and negative work decreases it.
• Algebraically, W = DE    or   F D = D(KE + PE).
• Work done by an object reduces its total energy/
• Work will cause a change in potential energy, kinetic energy, or both.

Questions:

1.  An arrow is placed on a bow, the bowstring is pulled back, the arrow is shot straight up
     into the air, the arrow then comes back and sticks into the ground.  Describe all of the
     occurences and changes in work, PE, and KE.

2.  If it takes an average force of 5.5 N to push a 4.5 g dart 6.0 cm into a dart gun, how fast
     should the dart exit the gun?

3.  A 75 Kg person pushes on the ground through a distance of 30.0 cm to jump straight up.
     If they go 45 cm into the air, how much force did their legs exert on the ground?

4.  In running a sprint a 65 Kg athlete reaches their top speed of 8.5 m/s after 10 m.
     What was the average force they exerted on the ground.

5.  Bolo, the human cannon ball, is projected from a 3.5 m long barrel.  If Bolo (m=80 Kg) has
     a speed of 12 m/s at the top of his trajectory, 15 m above the ground, what was the average
     force on him while being "fired"?
 

Conservation of Energy           (notes that apply)    (solutions)

Concepts

• The conservation of energy refers to the total energy of a system remaining constant in the absence of any outside forces.
• The values of the potential and kinetic energies of the objects in the system can change, but the total energy of the system stays the same.
• In most cases a change in potential energy results in the opposite change in the kinetic energy.
• If energy does appear to leave the system, it usually exists as heat, light, or sound, but the total of all energies still remains the same.

Questions

1.  If you jump off of a table onto the floor, is your energy conserved?  If not, where does it go?

2.  A car moving down the road smashes into a tree.  Is the energy of the car conserved?
     If not, where does it go?

3.  If you are at the top of a 40 m tall toboggan run, how fast should you be going at the bottom?

4.  Does the steepness of the run in #3 affect how fast you will go at the bottom?

5.  A high jumper approaches the bar at 9.0 m/s.  What is the highest the jumper go if they do
    not use any additional push off the ground and they are moving at 7.0 m/s as they go over
    the bar?

6.  A roller coaster train is moving at 2.0 m/s at the top of the first hill (h=40 m).
     How fast should the train be moving at the top of a later hill which is 15 m high?

7.  You are on a swing which has a 4.0 m long chain.  If your maximum displacement from the
     vertical is 35°, how fast will you be moving at the bottom of the arc?
 

Elasticity                               (notes that apply)           (solutions)

Concepts

• Elasticity refers to the amount of kinetic energy conserved in an interraction or collision.
• In a perfectly elastic collision all of the kinetic energy is conserved.
• In a perfectly inelastic collision the two object stick together and move as one unit.
• Usually, the harder the objects involved in the collision, the more elastic the collision.
• Any kinetic energy which is "lost" shows up as heat, sound, light, or the deformation of the objects.
• In any collision, momentum is still conserved.

Questions

1. A 1000 Kg blue car moving at 25 m/s collides head on with a 1200 Kg red car moving at -15 m/s.  After the collision the two cars stick together.

a)  Use the momentum concepts to find the speed of the cars after the collision.
b)  What percentage of kinetic energy was "lost"?

2.  A 200 g ball is dropped from a height of 2.0 m.  If it loses 20% of its energy in the bounce
     off the floor, how high will it reach after the bounce?

3.  The height of the top hill of a roller coaster is 35.0 m.  If the train loses 35% of its energy by
     the time it gets to a hill which has a height of 20.0 m,

a)  does it have enough energy to make it over the first hill?
b)  if it does, how fast will it be moving as it goes over the second hill?

Work and Power                (back to the questions)

1.  None. The bag has no displacement.

2.  W = F D = 20.0 N (4.0 m) = 80 J

3.  Since the pull is not along the displacement you must first calculate the horizontal component
     of the 25.0 N force.
                                   F_h = 25.0 N(cos 30°) = 22 N
           W = F_h D = 22 N (25 m) = 550 J

4.  a)  your work = your force x displacement = 200 N (4.0 m) = 800 J
     b)  frictional work = frictional force x displacement = -150 N (4.0 m) = -600 J
     c)  work by gravity = 0 since F_g is perpendicular to the motion
     d)  net work = net force x displacement = 50 N (4.0 m) = 200 J  (notice that this is also the
          sum of your work and the frictional work)

5.  P = W / T = F D / T = 1000 N (30 m) / 20 s = 1500 W

6.  P = W / T = F D / T         F = P T / D = 1000 W (1.5 s) / 8.0 m = 188 N
 

Potential and Kinetic Energy          (back to the questions)

1.  PE = mgh = 2.0 Kg (9.8 m/s²)(1.5 m) = 29.4 J

2.  PE = mgh        h = PE / m g = 500 J / (40.0 Kg x 9.8 m/s²) = 1.3 m

3.  Since PE is proportional to h, if the h doubles so does the PE.

4.  KE = 1/2 mV² = 1/2 (1200 Kg)(15 m/s)² = 1.35 E5 J

5.  Since KE is proportional to V², if the V doubles then the KE increases by a factor of 4.

6.  KE = 1/2 m V²       V = sqrt(2KE / m) = sqrt(2 x 14400 J / 200 Kg) = 12 m/s

7.  TE = KE + PE = 1/2 m V² + m g h = 1/2 (1.5 Kg)(20 m/s)² + 1.5 Kg (9.8 m/s²)(15 m)
          = 300 J + 221 J
          = 521 J
 

Change in Energy               (back to the questions)

1.

1. You do work on the bowstring.
2. The bowstring has increased potential energy.
3. The bowstring loses potential energy and does work on the arrow.
4. The kinetic energy of the arrow is increased.
5. The arrow rises in the air, transferring kinetic energy into potential energy.
6. The arrow stops rising when its kinetic energy is zero and its total energy is all potential.
7. The arrow falls, transferring potential energy into kinetic.
8. The arrow loses kinteic energy when it does work when it hits the ground and does work on the dirt.
9. The temperature of the arrow is increased, the temperature of the dirt is increased, sound is created, and dirt is moved.
10. During the entire flight air resistance does negative work on the arrow, reducing its total energy.

2. W = DTE = DKE  since DPE = 0.       F D = KE_f - KE_o = KE_f  since KE_o = 0
                                                       5.5 N (0.06 m) = 1/2 (.0045 Kg) V_f²
                                                                    12 m/s = V_f

3.  W = DTE = DPE  since DKE = 0.       F D = PE_f - PE_o = PE_f  since PE_o = 0
                                                         F (0.3 m) = 75 Kg (9.8 m/s²)(0.45 m)
                                                                     F = 1103 N

4.  W = DTE = DKE  since DPE = 0.       F D = KE_f - KE_o = KE_f  since KE_o = 0
                                                         F ( 10.0 m) = 1/2 (65 Kg)(8.5 m/s)²
                                                                       F = 235 N

5.  W = DTE = D(KE + PE) = 1/2 mV² + m g h
                                          = 1/2 (80 Kg)(12 m/s)² + 80 Kg(9.8 m/s²)(15 m)
                                      W = 17520 J
     since W = F D     F = W / D = 17520 J / 35 m = 2000 N
 

Conservation of Energy                (back to the questions)

1.  No.  It goes into increasing the temperature of your shoes and the floor as well as sound.

2.  No.  It goes into increasing the temperature of the car and the tree as well as sound.

3.  TE is conserved between the top of the hill and the bottom.
             TE_top = TE_bottom           (PE + KE)_top = (PE + KE)_bottom
                                               since KE_top = 0 and PE_bottom = 0
                                                             PE_top = KE_bottom
                                                            m g h = 1/2 m V²     (mass cancels out)
                                             9.8 m/s² (40 m) = 1/2 V²
                                                          28 m/s = V

4.  No.  It is the vertical displacement which determines the speed, not the path taken.

5.  TE is conserved between the running athlete and the top of the jump.
          TE_run = TE_top                  (PE + KE)_run = (PE + KE)_top
                                                             KE_run = (PE + KE)_top
                                                            1/2 m V² = m g h + 1/2 m V²     (mass cancels out)
                                                     1/2 (9.0 m/s)² = 9.8 m/s² (h) + 1/2 (7.0 m/s)²
                                                                 1.6 m = h

6.  TE is conserved between the top of the first hill and the top of the later hill.
           TE first = TE later                (PE + KE)first = (PE + KE)later
                                                    (mgh + 1/2mV2)first = (mgh + 1/2mV2)later
                                                                        mass cancels out
                                   9.8 m/s2 (40 m) + 1/2 (2.0 m/s)2 = 9.8 m/s2 (15 m) + 1/2 V2
                                                                           394 J = 147 J + 1/2 V2
                                                                         22 m/s = V

7.  The height of the swing falls is due to its elevation when displaced from the vertical.
           h = 4 m - X           where X = 4 m (cos 35°)
                                                 X = 3.3 m
     h = 4 m - 3.3 m
           h = 0.7 m

           Total energy is conserved during the swing.
                        TE_top = TE _bottom
             (PE + KE)_top = (PE + KE)_bottom
              KEtop = 0    and   PE_bottom = 0
                        PE_top = KE_bottom
                       m g h = 1/2 m V²        (the m cancels)
           9.8 m/s² (0.7 m) = 1/2 V²
                       3.7 m/s = V
 

Elasticity                (back to the questions)

1.  a)                                  m_bV_b + m_rV_r = m_rbV_rb'
         1000 Kg(25 m/s) + 1200 Kg(-15 m/s) = 2200 Kg (V_rb')
                                                    3.2 m/s = V_rb'

     b)  KE_before = KE_b + KE_r
                         = 1/2 mV_b² + 1/2 mV_r²
                         = 1/2(1000 Kg)(25 m/s)² + 1/2(1200 Kg)(-15 m/s)²
                         =  4.5 E5 J
          KE_after = KE_rb = 1/2 m V_rb² = 1/2(2200 Kg)(3.2 m/s)²
                       = 1.1 E4 J

          % "lost" = (4.5 E5 J - 1.1 E4 J) / 4.5 E5 J = 0.98        98 % "lost"

2.  Since it loses 20% of its energy on the bounce, It still has 80%.
                 E_after = 0.80 (E_before)
                 m g h_after = 0.80 mgh_before      (m and g cancel)
                        h_after = 0.80 h_before
                        h_after = 0.8(2.0 m)
                        h_after = 1.6 m

3.  a)  Since it loses 35% of its energy, it still has 65% of it.  It could not go over any hill which
         is more than 65% of the initial hill.
                      0.65 (35 m) = 22.8 m       The later hill is less than 22.8 m
                                                             so it makes it over the hill

     b)  E_later = 0.65(E_first)
           (PE + KE)_later = 0.65(PE +KE)_first
           since V_first is approximately zero, KE_first = 0
           (PE + KE)_later = 0.65 PE_first
           (mgh + 1/2 mV2)_later = 0.65 m g h_first     (the m cancels)
           (9.8 m/s2(20 m) + 1/2(V2) = 0.65 (9.8 m/s2)(35 m)
                                               V = 7.3 m/s

4.  Believe it or not, the steel ball.  The steel ball deforms less on impact, thus absorbing less
    energy.  Less energy absorbed means more energy retained in the collision - more elastic.