1. JR/2008 Conservation of Energy Conservation of Energy:- the principle states that:- ENERGY CANNOT BE CREATED OR DESTROYED This can be put to good use in the solution of engineering problems A typical example can be seen when considering free falling objects eg. If a smooth ball of mass m kg is held h metres above a fixed surface and released it will fall due to gravity g. In energy terms the initial phase is potential energy (PE) mgh as the ball falls the PE is transferred into kinetic energy (KE) ½ mv 2 If energy losses are ignored then the PE at the start = KE at the end when the ball hits the surface.

2. JR/2008 Conservation of Energy IfPE at the start = KE at the end thenmgh=½ mv 2 as mass m is constant the equation can be transposed for v final velocity Students should recall this equation from previous work on free fall. It must be noted that if energy losses are included this equation cannot provide an accurate solution

3. JR/2008 Conservation of Energy FromPE at the start = KE at the end andmgh=½ mv 2 The concept of energy to work translation can also be included eg. If a smooth ball of mass 3 kg is held 10 metres above a fixed absorbent surface and released it will fall due to gravity g. Find the average retarding force provided by the surface on the ball. Initial PE (mgh) = 3kg x 10m x 9.81 = 294.3 J The velocity v on impact = = 14 ms -1 But without losses the KE on impact = PE at start = 294.3 J If the ball penetrates a depth of 15mm into the surface then work is done As work = force x distance the 294.3 J of energy will be given up to work As the distance moved into the surface = 15mm = 0.015m Then the average retarding force on the ball can be found

4. JR/2008 Conservation of Energy From before the KE at impact = 294.3J. (Note this was also the initial PE) This is transferred into work (Force x distance) F.s Thus 294.3 J = Average force F x 0.015m Giving F = 294.3/0.015 = 19620 N = 19.62 kN This will be the average force on the ball provided by the surface.

5. JR/2008 Alternative solution: From the free fall equation it was noted that velocity at impact was 14 ms -1 Treating the problem as a simple motion task: u = 14 ms -1 v = 0 ms -1 (when the ball finally stops 15mm into the surface) a = ? s = 15mm = 0.015m t = ? to find acceleration (retardation) a use acceleration == -6533.33 ms -2 (retardation) To find the retarding force F use N2; F = ma = 3kg x 6533.33 ms -2 = 19.6 kN Review both methods and consider which you feel to be the most efficient method Conservation of Energy

6. JR/2008 Conservation of Energy Try the following: A bullet of mass 10 grams leaves a gun at a velocity of 350 ms -1. If the bullet hits a stationary target which provides an average retarding force of 1234 N calculate how far into the target the bullet will penetrate before stopping. Neglect air resistance on the bullet. Soln.Initial KE = KE at impact (no losses) = ½ mv 2 when mass m = 10 gram = 0.01kg, velocity v = 350 ms -1 KE = ½ x 0.01 x 350 2 = 612.5J work done on target = 612.5J work = Force x distance (Fs) When retarding force F = 1234N Distance moved = 612.5 J ÷ 1234N = 0.496metres = 500 mm penetration Consider a bullet proof vest !!!