Conservation of Energy By Chuanqi Zhang Rong Zhu

The Basics of Conservation -The conservation of energy states that in a closed system, the amount of energy will remain constant over time. -Energy will change from one form to another, but the total energy in the system will not change. Energy cannot be created nor destroyed. -Conservation of energy applies in many situations, such as elastics, pendulums, and simply letting an object fall.

Transfer of Kinetic to Potential -An object above ground level has an initial potential energy. As the object falls, its potential energy decreases and its kinetic energy increases. At the bottom of its fall, its kinetic energy is equal to the initial potential energy. PE=KE, or mg=1/2mv^2

Pendulums -The pendulum is a simple demonstration of conservation of energy. -At the top of the pendulum, it has maximum potential energy and no kinetic energy. -At the bottom, it has maximum kinetic energy and no potential energy. -Once it goes back to the top, it once again has maximum potential energy and no kinetic energy

Elastics -Conservation of energy can also be found in elastics. -The amount of potential energy stored is equal to the work done in stretching/compressing the spring. -As the spring returns to its original position, work is done. PEs=1/2kx^2=W

Work-Energy Relationship -In real situations, there is air resistance or friction. -When friction or air resistance is present, then an object in motion or with kinetic energy will convert some of that energy into heat or internal energy, Q since it is doing work against friction. -The law of conservation of energy means the sum of kinetic, potential and internal energy is constant. Total energy=PE+KE+Q

Quiz Time!

Question 1 A ball is dropped from 10 meters above the ground. The ball has a mass of 2kg. Using conservation of energy, calculate how fast it will be traveling when it hits the ground. Note: neglect air resistance.

Solution As the ball falls, energy is conserved. This means that its potential energy gets converted into kinetic energy. PE = KE mgh=1/2 mv2 Plug in the values (2kg)(9.81m/s2)(10m)=1/2(2kg)v2 v=14 m/s

Question 2 A toy gun that shoots rubber balls of mass m=0.2 kg is loaded by inserting the ball into the barrel of the gun. The spring inside the gun has a spring constant of k= 100 N/m. When the gun is loaded, the spring is compressed by an amount x= 0.05 m. The gun is pointed straight up. How far up will the rubber ball go?

Solution PE (spring)=1/2kx2 When the trigger is pulled, the spring rapidly expands, releasing its potential energy into kinetic energy KE = 1/2 mv2 PE (spring) = KE 1/2 kx2=1/2 mv2 Plug in values and solve v=1.11m/s PE = KE = mgh=1/2 mv2 h=0.0628m

Question 3 A block of mass m= 10 kg can slide along a curved track The block, initially at a height h= 3 meters above the ground is released. It slides down the frictionless track, and hits the spring- stopper at the end of the track. The spring has a spring constant k= 50N/m. How far does the spring compress when it brings the block to a stop?

Solution When the block slid down the curve and onto the flat section of the track, it hits the spring and the spring compresses until the block is stopped PE = PE (spring) mgh=1/2 kx2 (10kg)(9.81m/s2)(3m)=1/2(50N/m)x2 x = 3.42 meters

Question 4 A ball is thrown up vertically with a velocity of 20m/s. What height does it reach, assuming there is no friction?

Solution PE=KE mgh=1/2 mv2 since m is on both side of the equation, they will cancel out. You don't even the mass to find the solution! gh = 1/2 v2 (9.81m/s2)h = 1/2 (20.0m/s)2 h = 20.4 meters

Question 5 A 1.0 kg pendulum bob is released from a height of 1 m above the base level. Assuming there is no friction or air resistance, what is the potential energy of the bob at the point of release? What is the velocity of the bob at the bottom of its swing?

Solution PE = mgh PE = (1.0kg)(9.81m/s2)(1m)= 9.81 J KE (on the bottom) = PE (on the top) KE = 1/2 mv2 (9.81J) = 1/2 (1kg) v2 v = 4.43 m/s