The law of Conservation Energy

Definition: the law of Conservation of Energy is the total energy of an isolated system which remains constant from the beginning to the end.  

                                                 Abbreviated Version:

                                                       Ei = Ef

                                                 Extended Version:

                              E​Ki​​+U​gi​​+U​si​​ = E​Kf​​+U​gf​​+U​sf​​+E​Hf​​ or
mvі^2/2 + mgh = mvf^2/2 + mgh

                                                SI units of Energy:

                                                       Joule (J)

                                                    American Units:  

                                              Calories

                                                 BTU

Conservation of energy tells us that the sum of the total kinetic energy and potential energy not change. The video point out that the train of coaster cars speeds up as they lose height. Thus, their original potential energy is transformed into kinetic energy. As the ride continues, the train of cars are continuously losing and gaining height.

Problem 1:

Aaqib rides his skateboard at a local skate park. He starts from rest at the top of the track as seen in the figure below and begins a descent down the track, always maintaining contact with the surface. The mass of the skateboard is negligible, as is friction except where noted.

(a) What is Aaqib’s speed when he reaches the bottom of the initial dip, 18.0 m below the starting point?

(b) He then ascends the other side of the dip to the top of a hill, 8.0 m above the ground. What is his speed when he reaches this point?

Solution:

a. From the conservation of energy: Potential energy at the top of the 18 m transforms into the Kinetic energy at the bottom of the dip.

Known,

m = 66kg

h = 18m

g = 9.81

Unknown,

Vi = ?

b. From the conservation of energy: Potential energy at the top of the 18 m transforms into the Kinetic and Potential energy at the top of a hill.

Known,

m = 66 kg

h1 = 18 m

h2 = 8 m

Unknown,

V2 = ?

Answer and     

               

Problem: 2

Aayan starts biking from rest with mass of 65kg, how fast is he need to move to reach 25m high hill ?

Known,

m = 65 kg

h = 25 m

Unknown,

V = ?

                       2mgh = 1/2mv2

                           2gh = v2

                                   √2gh  = v

            √2(9.8)925)= v

             22.13 m/s = v

The maximum height is 8 meters. The initial velocity of the skater is 0 ; because he started from rest where his potential gravitational energy stored up but there is no kinetic energy.

vi = 0

The skater placed at the height of 4.8 meters, which will take him to the peak of the loop. Kinetic energy will help the skater make it down the rest of the loop.

mvі^2/2 + mgh = mvf^2/2 + mgh

gh= gh mv^2/2

(9.8)(7) - (9.8)(4.8) = v^2

21.56 = v^2

4.64 m/s = v

                                   

​

The skater's total energy does not change  as he moves down the ramp. Potential energy has been decreased at the bottom of the ramp, and is converted to kinetic energy.

mvі^2/2 + mgh = mvf^2/2 + mgh

(9.8)(8) = mvf^2/2

78.4 (2) = (mvf^2/2)/2

156.8 = vf^2

12.52 m/s = vf