Conservation+of+Energy

=Conservation of Energy=

**Introduction**
The law of conservation of energy states that energy can not be gained or lost, created or destroyed. Energy can only be converted from one form to another(potential or kinetic). These principles form the idea that energy is conserved. Energy=Energy, E=E The basic understanding; you can not make something out of nothing is the same concept with the law of conservation of energy. Mass can not be created or destroyed therefore the idea that mass equals energy is formed

__**Energy**__ Energy is one of the most important concepts used in every day life with a variety of forms. We define energy as work over time. Since power equals energy, energy equals work over time.

__**Work**__
Work can be described as the displacement of an object due to the force applied on it. Work equals forces times displacement.

Work will be doubled if either force or displacement is doubled.

__**Forms of Energy**__
Energy can be formed into kinetic energy, potential, mechanical, chemical, heat and light.

__**Kinetic Energy**__
Kinetic energy is energy in motion doing work on another object. m= mass of the object v= velocity of the object

__**Potential Energy**__
Potential energy can be described as an object with the potential to do work.

The equation for gravitational potential energy is: m=mass g=gravity constant ( ~10 m/s^2) h= height of the object

__**Forces**__
There are two general kinds of forces, conservative and nonconservative. A well-known conservative force would be gravity. Nonconservative force randomly disperse the energy that it holds forming heat or sound.

**Examples**
If the the ball rolls, without external forces like friction or air resistance, from point A to point D, the potential and kinetic energies __because__
 * __Hill Example Question 1__**

Answer: remain the same ; energy is neither created or destroyed.

With the conservation of energy equation, you can find the velocity of a moving object.

__**Spring Example Question 2**__ Intial length of spring = 1 meter Mass of block = 2 kg Spring constant (k)= 155.9 If a person attaches the block to the spring and pulls so that L=2 meters and releases it, what is the speed the block when it returns to it's initial length?

Answer: __Look at the bottom of the page__

__**Example Question 3**__ A 95 kg girl is running with a speed of 3 m/s. How much kinetic energy does she have? She grabs on to a rope that is hanging from the ceiling, and swings from the end of the rope. How high off the ground will she swing? Answer: Bottom of page

A skater skates down a frictionless hill of 100 meters, the ascends another hill, of height 90 meters, as shown in the figure below. What is the speed of the skater when it reaches the top of the second hill?
 * __Example Question 4__**

__**Example Question 5**__ A particle, under the influence of a conservative force, completes a circular path. What can be said about the change in potential energy of the particle after this journey? __**Example Question 6**__ A pendulum swings back and forth, the forces acting on the suspended object is gravity, the tension on the cord, and also air resistance. a) Which of the forces does no work? b) Which of these forces does negative work the whole time during the pendulum's motion?  __**Example Question 7**__ A 5.00 kg steel ball is dropped onto a copper plate from a height of 10.0 m. IF the ball leaves a dent 3.20 mm deep in the plate, what is the average force exerted by the plate on the ball during the impact?

__**Example Question 8**__ A projectile is launched with a speed of 40 m/s at an angle of 60 degrees above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.

__**Example Question 9**__ 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is above the bar is 1.0 m/s. Determine her altitude as she crosses the bar.

__**Example Question 10**__ A child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. If the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?

__**Answers:**__ Hill question 1- remain the same ; energy is neither created or destroyed. Spring question 2- __When the spring is stretched the potential energy equals to the gravitational potential energy. Since the total energy is conserved. the energy that the initial length is equal to stretched length. We can end up to the equation__

__(½)mv² =(½)kx² -mgx __

__(1/2)(2)(v)^2=(1/2)(155.9)(1)^2- (2)(10)(1)__ __(1)(v)^2= 77.95 - 20__ __(v)^2=57.95__ __v ~ 7.6 m/s__ Question 3- Using **Use ∆Ep = ∆Ek** First find kinetic energy: 1/2mv^2 (1/2)95(3)^2 = 427.5 J

To find out high she will swing: 427.5=mgh 427.5= 95(10)(h) 427.5=950(h) h= .45m Questions 4- W=Fx=mgh=10mg //1/2mv^2= 10mg// cancel the mass and solve for v  v=20^-2g=14 m/s Question 5- Potential energy is the energy of configuration of a system. If the particle returns to its initial position, the configuration of the system is the same, and must have the same potential energy. Question 6- a) The tension in the cord does no work. Motion on the pendulum is always perpendicular to the cord. b) The air resistance does negative work at all time because the air resistance is going to be acting on it on the opposite direction. Question 7- Mass = 5.00 kg Height = 10.0 m Dent depth = 3.20 mm Using the for F(average) = (mgh)/ d  = (5.0 x 9.8 x 10.0)/ 0.32 x 10^-2 = 1.53 x 10^5 N Upward Question 8- Initial speed=40 m/s Angle of Inclination=60 degrees Maximum height=h (Position A) Potential Energy=0 Kinetic Energy of the projectile at position A= (1/2)mv^2= 1/2m(40)^2=800mJoules (Position B) Potential energy=mgh Kinetic Energy of the projectile at position B= (1/2)m(vcos60 degrees)^2= 200m J 800 - 200 = 9.8h h = 61.2m Question 9- Mass of vaulter=50 kg Initial Kinetic Energy= (1/2)mv^2=1/2 x 50 x 10 x 10= 2500 J Speed of Vaulter= 10m/s Final Kinetic Energy= (1/2) x 50 x (1.0)^2= 25J 2500= 25+50x9.8xH H=5.05m Question 10- Mass= 50.0 kg Initial speed= 0 Speed of sled at the bottom of the hill= 3.00 m/s H= (v^2)/2g =[(3.00 m/s)^2] / 2x(9.8 m/s^2) =.46 m