Everything Science Grade 10

CHAPTER 22. MECHANICAL ENERGY 22.2 EP= mgh = (1)(9,8)(2) = 19,6 J EP= mgh = (1)(9,8)(0) = 0 J The potential energy is a maximum. ...

22.2 CHAPTER 22. MECHANICAL ENERGY EP = mgh = (1kg)(9, 8 m·s−^2 )(4m) = 39, 2 J Example 2: More gravitational potential energy Q ...

CHAPTER 22. MECHANICAL ENERGY 22.2 EP=mgh First we need to calculateh. The height of the ball above the ground when the girl sho ...

22.3 CHAPTER 22. MECHANICAL ENERGY a. How much potential energy did the boy gain by climbing onto the roof? b. The boy now jumps ...

CHAPTER 22. MECHANICAL ENERGY 22.3 whereEKis the kinetic energy (measured in joules, J) m = mass of the the object (measured in ...

22.3 CHAPTER 22. MECHANICAL ENERGY Example 3: Calculation of Kinetic Energy QUESTION A 1 kg brick falls off a 4 m high roof. It ...

CHAPTER 22. MECHANICAL ENERGY 22.3 Example 4: Kinetic energy of 2 moving objects QUESTION A herder is herding his sheep into the ...

22.3 CHAPTER 22. MECHANICAL ENERGY higher mass. Checking units ESAHM According to the equation for kinetic energy, the unit shou ...

CHAPTER 22. MECHANICAL ENERGY 22.3 We are given the mass of the bulletm= 150 g. This is not the unit we want mass to be in. We ...

22.4 CHAPTER 22. MECHANICAL ENERGY More practice video solutions or help at http://www.everythingscience.co.za (1.) 00ar (2.) 00 ...

CHAPTER 22. MECHANICAL ENERGY 22.5 Step 1:Analyse the question to determine what information is provided The ball has a massm= ...

22.5 CHAPTER 22. MECHANICAL ENERGY DEFINITION: Conservation of mechanical energy Law of Conservation of Mechanical Energy: The t ...

CHAPTER 22. MECHANICAL ENERGY 22.5 EM 1 = EM 2 EP 1 +EK 1 = EP 2 +EK 2 mgh+^12 mv^2 = mgh+^12 mv^2 (1kg)(9, 8 m·s−^1 )(2m) + 0 = ...

22.5 CHAPTER 22. MECHANICAL ENERGY What is the velocity (i.e. fast, slow, not moving) of the marble when it reaches the other e ...

CHAPTER 22. MECHANICAL ENERGY 22.5 Example 7: Using the Law of Conservation of Mechanical Energy QUESTION During a flood a tree ...

22.5 CHAPTER 22. MECHANICAL ENERGY At the bottom of the waterfall,h= 0m, soEP 2 = 0. ThereforeEP 1 =EK 2 or in words: The kineti ...

CHAPTER 22. MECHANICAL ENERGY 22.5 SOLUTION Step 1:Analyse the question to determine what information is provided The mass of t ...

22.5 CHAPTER 22. MECHANICAL ENERGY ciples”: (v 2 )^2 = 2gh 1 (v 2 )^2 = (2)(9. 8 m·s−^1 )(0, 5 m) (v 2 )^2 = 9, 8 m·s−^1 ·m v 2 ...

CHAPTER 22. MECHANICAL ENERGY 22.5 1. the velocity of the roller coaster when it reaches the top of the loop 2. the velocity of ...

22.5 CHAPTER 22. MECHANICAL ENERGY Step 4:Calculate the velocity at the bottom of the loop Again we can use the conservation of ...