6.1 Work

Work done on a body is defined by DW = F dot DL = F cosq DL where q is the angle between the force vector F and the distance vector DL over which the force acts.
It takes energy to do work. Work is done by consuming energy or transferring energy. The amount of work done is measured by the amount of energy transferred or transformed. It should be self evident that energy is required to get work done, while work transfers energy.

In the MKS units, work done is measured in units of Joule.

     1 J = 1 N x 1 m = 0.5 N x 2 m = 10 N x 0.1 m, ....... etc

Work can transfer into energy; energy can transfer into work.

The kinetic energy can be calculated by calculating the work done to change the velocity of a body. The potential energy can be calculated by calculating the work done to change the elevation of a body relative to a chosen reference level (the sea level for example).

6.2  Kinetic Energy

Assuming the frictional force can be turned off in Figure 1. The work done on the body by the horizontal component of the force, F cosq, goes to accelerating the body of mass m. Acceleration leads to change of velocity.

The enegy associated with a moving object is called the kinetic energy given by

        KE = (1/2) mV**2       DKE = (1/2)mVf**2 - (1/2)mVi**2

Derive the above expression for the kinetic energy. The kinetic enegy is derived by calculating the work done by a force leading to changing the velocity as shown in Figure 1 (b):

   dW = (Fcosq) dL = (ma) dL = m (dV/dt) dL= m dV/dt (dL/dt) dt= mV (dV/dt) dt = m VdV

Integrating both sides of the equation leads to   W = (1/2) m V**2 = KE

If you are not comfortable with integration, repeat the above derivation using the finite difference notation leading to

DW = mVav DV = m (1/2)(Vf + Vi) (Vf - Vi) = (1/2) mVf**2 - (1/2) mVi**2 = DKE

The force accelerates the body to increase its velocity. The work done by the force goes into increase the kinetic energy of the moving body.

Example: A 100 kg runner improved his 100 meter dash record from 11 s to 9.8 s to qualify for the Olympic. How much work must be done by the runner on the track to shave off 1.2 seconds?

The average speed before the Olympic trial is Vb = 100 m/11 s = 9.1 m/s. The average speed at the Olympic trial is Va = 100/9.8 = 10.2 m/s

   DW = DKE = (1/2)mVa**2 - (1/2)mVb**2 = (100 kg/2) (10.2**2 - 9.1**2)
          = 50 kg x (104 - 82.8) m/s**2 = 1,060 J

6.3 Potential Energy

6.3.1 If g is constant

The work done on the body to overcome the gravitational force leads to increase of the potential energy given by

        PE = mg h       DPE = mg Dh

where Dh is the increase in altitude relative to a reference height level.   The potential energy can be derived by calculating the work done by an external force against the gravitational force, as shown in Figure 1(c). If the force in Figure 1 (c) lifts up the block and increase its height by Dh, the work done on the block can be calculated by

          DW = F dot DL = (Fsinq) Dh = mg Dh = DPE (in Joules) ---------------- (A)

Since g is constant, (A) can be readily integrated once to yield  PE = mgh.

Example: A 10 kg mass is lifted up from the ground and placed on a 2-meter stand. How much work is done on the mass?

The work is done to overcome the gravitational force. The energy is transferred and stored as the potential enegy of the body after having been elavated by 2 meters.

W = mg x Dh = 10 (kg) x 9.8 (m/s**2) x 2 (m) = 98 (N) x 2 (m) = 196 J

6.3.2 If g is not constant

The gravitational acceleration g = 9.8 m/s**2 =constant only near the earth surface. Chapter 5 showed that Fg = G m M/r**2 so that g(r) = GM/r**2.

On the ground, g(r=1Re) = GM/Re**2 = 9.8 (m/s**2) so that GM = 9.8 x Re**2

Therefore, g(R) = 9.8 x Re**2/r**2 = 9.8/R**2 where R = (r/Re) in units of earth's radius Re = 6371 km so that R = 1 at the sea level, R = 2 if an object is 1 Re above the sea level.

Equation (A) can be rewritten in differential form:

    dPE = m g dh = (m 9.8/R**2) dr = [m g(1)/R**2] Re d(r/Re)

          = [m g(1)/R**2] RedR = [m g(1)Re]dR/R**2

If you have had calculus, you can integrate the above equation once to obtain

         PEG = - (m 9.8 Re /R) -------------------------------------------------------- (B)

If you have not had calculus, you follow the finite difference method:

     DR = Rf - Ri ;   Rav = (Rf + Ri)/2 = (2Ri + DR)/2

     DPE = [m g(1) Re] DR/Rav**2 = [m g(1) Re] (Rf - Ri)/[(2Ri + DR)/2]**2

     [(2Ri + DR)/2]**2 = (4Ri**2 + 4Ri DR + DR**2)/4
                               = Ri(Ri + DR) = Ri Rf     

      DR**2 can be neglected because (DR)**2 << Ri DR as DR appraoches zero

      DPE = [m g(1) Re] (Rf - Ri)/Ri Rf = m g(1) Re (1/Rf - 1/Ri)

      PEG = - m g(1) Re/R  ----- if PEG = 0 is chosen at R = infinity.

6.4.1 Conservation of Total Energy        

Conservation of Total Energy means that the total energy of a system is constant if no external force is doing work on the system. This is proved as follows:
By definition:       DW = F dot DL    and      DW = DET
If external force F = 0, it follows that no work is done on the system:
       DW = DET = 0
       ET = KE + PE + Heat Energy + Electrical Energy + ...... = Constant ------------ (A)

6.4.2 Conservation of Mechanicle Energy (A Special Case)

In a system where only gravitational force in involved, we have shown that the work done by the gravitational force will go into mechanical energy defined as the sume of potential energy and kinetic energy, i.e., E = PE + KE, no other forms of energy will be produced.

Now, if the gravitational force is taken as a part of the system. then there is no external forces of any kinds acting on the system. Therefore, (1) is reduced to

          ET = E = KE + PE = Constant ---------------------------------------------------- (B)

For a pendulum, (B) becomes

6.5 Applying Conservation of Mechanical Energy

Ex 1: If the maximum height a pendulum swings is Dhmax = 10 cm, what is the maximum speed of the pendulum? Conservation of mechanical energy in (1) leads to
        m.g.Dhmax = (1/2)m.(Vmax)**2 so that Vmax = 1.4 m/s

Ex 2: The pole vaulter in example 6.9 clears 7 m bar. The vaulter's center of gravity is at 1 m (meaning Dhmax = 6 m). How fast the vaulter must run to clear the bar? (i.e., the take off speed, Vmax  = ?). It is interesting to note that the mass of the vaulter drops out of the equation.  
Dhmax = 6 m so that Vmax  = (2gDhmax)**0.5 = 10.8 m/s that is Olympic 100 m gold speed.

The escape velocity from earth's gravity 

From the conservation of energy, KE + PE = constant, one obtains

       KE - m x 9.8 x Re / R = constant -----------------------------------------------(C)

To escape from earth's gravity requires: Rf = infinite, and KEf = 0
(C) is reduced to KEf - m x g(1) x Re /Rf = 0 or (1/2) m (Vesc)**2 = m x 9.8 x Re /1  
   
     Vesc = (2 x 9.8 x 6.371 x10**6)**0.5 = 11.2 km/s

Ex:
(A) Show that the geocentric radius R of a satellite moving at a speed Vo in a circular orbit is given by  R = 9.8 Re/Vo**2

(B) Show that the escape speed Vesc of a satellite in a circular orbit of geocentric radius R is given by Vesc = Vo (2)**(1/2)

If R = 4 (i.e., 3 Re above the Earth surface), then Vo = (9.8 Re/R)**(1/2) = 3.16 km/s and Vesc =  Vo (2)**(1/2) = 4.46 km/s.

6.6 Power: Power is defined as the rate of change of work or energy transfer  

                       P = dW/dt     (in units of Watts ---- 1 W = 1 J / 1 s) -----------(D)

Substituting dW = F dot dS and dV = dS/dt into (D) yields: P = F . V = F u
where u is the velocity component parallel to the force vector F.

Ex: A rocket engine produces 890 N thrust moving at an average speed of 314 m/s for 20 s. What is the average power delivered by the engine and the total enegy transferred from the engine to the rocket during the burn?

P = 890 N x 314 m/s = 279 KW; total energy delivered E = 279 KW x 20 s = 5,580 KJ

HW Chap 6: P45, P 51, P73, P79