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Potential Energy and Kinetic Energy.  Potential Energy = -Work by conservative forces  Wnc = Delta KE + Delta PE  KE=.5 mv^2  PE= mg delta H.

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Presentation on theme: "Potential Energy and Kinetic Energy.  Potential Energy = -Work by conservative forces  Wnc = Delta KE + Delta PE  KE=.5 mv^2  PE= mg delta H."— Presentation transcript:

1 Potential Energy and Kinetic Energy

2  Potential Energy = -Work by conservative forces  Wnc = Delta KE + Delta PE  KE=.5 mv^2  PE= mg delta H

3  A 100kg bus travels on a hilly road. The speed of the bus is 26.82 m/s. The bus is coasting. Friction and air resistance are negligible. If the bus starts at an altitude of 55 m and goes down to 50m, what is the speed?.

4  V = 28.59 m/s  Ei = Ef because all forces are conservative  E= Ke+Pe=.5mv^2 + mgh .5mvi^2 + mghi =.5mvf^2 + mghf

5  Four identical balls are thrown from the top of a cliff, each with the same speed. The first is thrown straight up, the second is thrown at 30° above the horizontal, the third at 30° below he horizontal, and the fourth straight down. How do the speeds and kinetic energies of the balls compare as they strike the ground…

6  All speeds and KE finals are the same  All are at the same speed a the cliff peak and have same mass  Thus Kei=.5mv^2  All are the same at the start Pe = mgh all start from with the same velocity from the same height, thus delta H, from time zero and time final, is the same. All started with the same PE then

7  A 45 kg box is pushed up a 21 m ramp at a uniform speed. The top of the ramp is 3.0 m higher than the bottom.  What is the potential energy of the box at the top of the ramp relative to the bottom of the ramp?

8  PE = -mgh final – -mgh initial  PE = -45(-9.81)(3) – 45(- 9.81)(0)  = 1324.35 Joules

9  What work was done pushing the box up the ramp if friction were negligible and v final was 4 m/s?

10  Work = delta KE + delta PE  Delta KE =.5mvf^2 -. 5mvi^2 .5(45)(4)^2 -.5(45)(o)^2  =360 Joules  PE= -mgh final - - mgh initial  -(45)(9.81)(3) – -(45)(9.81)(0)  = 1324.35 Joules  Work = 360+1324.35 = 1684.35 Joules

11  Assume the ball weighs 1 kg

12 A. 50J B. 10 m/s C. 50J D. 50J E. 10 m/s F. 0 J G. 100 J H. 22.4 m/s I. 0 J J. 100 J K. 22.4 m/s


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