CK-12 Physics Concepts - Intermediate Answer Key Chapter 9: Energy 9.1 Kinetic Energy Practice Questions 1. 2. 3. Potential energy is present in objects that are ____________. Kinetic energy is present in objects that are ___________. What formula is given for kinetic energy? Answers 1. 2. 3. Stationary. Moving. KE = (1/2) mv^2 Review Questions 1. 2. A comet with a mass of 7.85 × 1011 kg is moving with a velocity of 25,000 m/s. Calculate its kinetic energy. A rifle can shoot a 4.00 g bullet at a speed of 998 m/s. a. Find the kinetic energy of the bullet. b. What work is done on the bullet if it starts from rest? c. If the work is done over a distance of 0.75 m, what is the average force on the bullet? d. If the bullet comes to rest after penetrating 1.50 cm into a piece of metal, what is the magnitude of the force bringing it to rest? Answers 1 1 π 1. Using πΎπΈ = 2 ππ£ 2 βΆ πΎπΈ = 2 (7.85 × 1011 ππ) (25,0002 π ) βΆ The comet’s kinetic energy is 2.45 × 1020 J. 2. 1 π a. Using πΎπΈ = 2 ππ£ 2 βΆ (. 004ππ) (9982 π ) = 1992 J. (Remember to convert 4.00 g to kg.) b. ππππ‘ = βπΎπΈ βΆ 1992 J. π€ 1992π½ c. Using π = πΉ βΆ .75π = 2656 N. 1 d. Using π€ π =πΉβΆ 1992π½ 0.015π = 132,800N. 9.2 Potential Energy Practice Questions 1. 2. 3. What is the definition of energy? Name two types of potential energy. How is energy transferred from one object to another? Answers 1. 2. 3. Energy is the ability or capacity to move an object (to do work). Chemical, Gravitational, Elastic, Electrical, and Nuclear. Energy is transferred from one object to another by doing work. Review Questions 1. A 90.0 kg man climbs hand over hand up a rope to a height of 9.47 m. How much potential energy does he have at the top? 2. A 50.0 kg shell was fired from a cannon at earth’s surface to a maximum height of 400m. a. What is the potential energy at maximum height? b. It then fell to a height of 100. m. What was the loss of PE as it fell? 3. A person weighing 645 N climbs up a ladder to a height of 4.55 m. a. What work does the person do? b. What is the increase in gravitational potential energy? c. Where does the energy come from to cause this increase in PE? Answers π 1. Using ππΈ = ππβ βΆ 90.0ππ ∗ 9.80 π 2 ∗ 9.47π βΆ 8350 J. 2. π a. Using ππΈ = ππβ βΆ 50.0ππ ∗ 9.80 π 2 ∗ 400π = 196,000 J. π b. Using ππΈ = ππβ βΆ 50.0ππ ∗ 9.80 π 2 ∗ 100π = 49,000π½ Subtract initial PE from final PE : 196,000 − 49,000 = 147000 J. 3. a. Using ππΈ = (ππ)β βΆ 645π ∗ 4.55π = 2930 J. b. Starts from 0: 2930 J. 2 c. The energy increase is the same as the work done against the gravitational force. The person at the top of the ladder has more distance to fall and thus more potential energy than he/she would at the bottom of the ladder. 9.3 Conservation of Energy Practice Questions 1. 2. 3. What happens when one ball is pulled up to one side and released and why? What happens when three balls are pulled up to one side and released and why? What happens when two balls are pulled out from each side and released and why? Answers 1. 2. 3. The potential energy of the ball is converted into kinetic energy as it falls. When it hits the four stationary balls, the kinetic energy is transferred through them to the last ball which rises to the same height as the starting ball, converting the kinetic energy back into potential energy. The process is then repeated in the opposite direction. When three balls are pulled up, the potential energy is again converted into kinetic energy. This time, the kinetic energy of the leftmost three balls is transferred into the rightmost three balls. The kinetic energy is converted back into potential energy and the process is repeated in the opposite direction. When two balls are pulled out from each side and released, they fall as before. The kinetic energy from both directions impacts the stationary balls at the same time. The central balls have no net force and so they remain stationary. The outer balls bounce back as if they had hit a solid wall and bounced off. They lose kinetic energy and gain potential energy as they rise from the impact point. Review Questions 1. A 15.0 kg chunk of ice falls off the top of an iceberg. If the chunk of ice falls 8.00 m to the surface of the water: a. what is the kinetic energy of the chunk of ice when its hits the water, and b. what is its velocity? 2. An 85.0 kg cart is rolling along a level road at 9.00 m/s. The cart encounters a hill and coasts up the hill. a. Assuming the movement is frictionless, at what vertical height will the cart come to rest? 3 3. 4. 5. 6. b. Do you need to know the mass of the cart to solve this problem? A circus performer swings down from a platform on a rope tied to the top of a tent in a pendulum-like swing. The performer’s feet touch the ground 9.00 m below where the rope is tied. How fast is the performer moving at the bottom of the arc? A skier starts from rest at the top of a 45.0 m high hill, coasts down into a valley, and continues up to the top of a 40.0 m high hill. Both hill heights are measured from the valley floor. Assume the skier puts no effort into the motion (always coasting) and there is no friction. a. How fast will the skier be moving on the valley floor? b. How fast will the skier be moving on the top of the 40.0 m hill? A 2.00 kg ball is thrown upward at some unknown angle from the top of a 20.0 m high building. If the initial magnitude of the velocity of the ball is 20.0 m/s, what is the magnitude of the final velocity when it strikes the ground? Ignore air resistance. If a 2.00 kg ball is thrown straight upward with a KE of 500 J, what maximum height will it reach? Neglect air resistance. Answers 1. π a. Using ππΈ = ππβ βΆ 15ππ ∗ 9.8 π 2 ∗ 8π βΆ 1180 J 2π 2(8π) b. First solve for time to fall using: π‘ = √ π βΆ √ π = 1.3π . Then solve for 9.8 π 2 π velocity using: π£ = ππ‘ βΆ 9.8 π 2 ∗ 1.3π =12.5 m/s 2. 1 π 2 1 a. Solve for KE using πΎπΈ = 2 ππ£ 2 βΆ 2 (85.0ππ) (9.00 π ) = 3442.5π½. Then solve for height using 3442.5J as PE with: ππΈ ππ 3442.5π½ = β βΆ 85ππ∗9.8 π/π 2 = 4.13 m b. No. To solve the problem, you equate the potential and kinetic energy equations: . Since mass is on both sides of the equation, it cancels out. Thus, we can determine the maximum height based only on the cart’s speed. 2π 2(9π) 3. First solve for time to fall using: π‘ = √ π βΆ √ π = 1.35π . Then solve for velocity 9.8 π 2 π using: π£ = ππ‘ βΆ 9.8 π 2 ∗ 1.35π =13.23 m/s The performer’s velocity is 13.2 m/s at the bottom of the swing. 4. 2π 2(45π) a. First solve for time to fall using: π‘ = √ π βΆ √ π π 9.8 2 π velocity using: π£ = ππ‘ βΆ 9.8 π 2 ∗ 3.03π =29.7 m/s 4 = 3.03π . Then solve for b. Speed at the top of hill#2 is equal to speed at the bottom – acceleration 2π π 2∗40π due to g over 40m. Acceleration over 40m = π ∗ √ π = 9.8 π 2 ∗ √ π π 9.8 2 π = 28 π . Speed at the bottom of the hill = 29.7 m/s, subtract 28 m/s to get speed at the top of hill#2: 1.7m/s 1 1 π 2 5. Solve for final KE using 2 ππ£ 2 + ππβ = 2 (2ππ) (20 π ) + π 2(792π½) (2ππ) (9.8 2 ) (20.0π) = 792π½. Then solve for final velocity using π£π = √ π 2.0ππ = √792 = 28.1 m/s. ππΈ 500 6. Using β = = = 25.51 m. ππ 2∗9.8 9.4 Elastic and Inelastic Collisions Practice Questions 1. Explain what happened in the first demonstration on elastic collisions. 2. Explain what happened in the second demonstration on inelastic collisions. 3. Assuming the first carts started at the same speed in both demonstrations, explain why the inelastic collision ended slower than the elastic collision. Answers 1. The momentum from the first cart was completely transferred into the second when the collision occurred and kinetic energy was unchanged. 2. The momentum from the first cart was shared between the two carts when the collision occurred and kinetic energy decreased 3. The end product of the inelastic collision had more mass than the elastic collision. Because momentum remained unchanged in both demonstrations but the mass of the second increased, the velocity had to decrease or else the momentum would also increase. Review Questions 1. A 4.00 kg metal cart is sitting at rest on a frictionless ice surface. Another metal cart whose mass is 1.00 kg is fired at the cart and strikes it in a one-dimensional elastic collision. If the original velocity of the second cart was 2.00 m/s, what are the velocities of the two carts after the collision? 2. Identify the following collisions as most likely elastic or most likely inelastic. a. A ball of modeling clay dropped on the floor. 5 b. c. d. e. A fender-bender automobile collision. A golf ball landing on the green. Two billiard balls colliding on a billiard table. A collision between two ball bearings. Answers 1. Solve for cart #1’s velocity using: π1 = 4ππ, π2 = 1ππ, π£π1 = 0 a. π£π1 = (π1 −π2 )π£π1 +2π2 π£π2 b. π£π2 = 2π1 π£π1 −(π1 −π2 )π£π2 π1 +π2 π1 +π2 βΆ βΆ π π (4ππ−1ππ)0 +2(1ππ)(2) 4ππ+1ππ π π π π (2∗4ππ∗0 )−(4ππ−1ππ)2 4ππ+1ππ = .8 π π , π£π2 = 2.00 π π π π = −π. π π π c. The first cart’s final velocity is 0.8 m/s, and the second cart’s final velocity is -1.2 m/s. 2. a. b. c. d. e. Inelastic – change in kinetic energy because the clay will compress. Inelastic – change in kinetic energy because both cars will compress. Inelastic – change in kinetic energy because the grass will compress. Elastic Elastic 6
