Download Intermediate Answer Key Chapter 9: Energy and more Exams Physics in PDF only on Docsity!
CK-12 Physics Concepts - Intermediate
Answer Key
Chapter 9: Energy
9.1 Kinetic Energy
Practice
Questions
- Potential energy is present in objects that are ____________.
- Kinetic energy is present in objects that are ___________.
- What formula is given for kinetic energy?
Answers
- Stationary.
- Moving.
- KE = (1/2) mv^
Review
Questions
- A comet with a mass of 7.85 × 10^11 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
- Using 𝐾𝐸 = 1 2 𝑚𝑣
2
( 7. 85 × 1011 𝑘𝑔) ( 25 , 0002 𝑚
𝑠 )^ ∶^ The comet’s kinetic energy is 2.45 × 10^20 J.
a. Using 𝐾𝐸 = 1 2 𝑚𝑣
𝑠 )^ =^ 1992 J.^ (Remember to convert 4.00 g to kg.) b. 𝑊𝑛𝑒𝑡 = ∆𝐾𝐸 ∶ 1992 J. c. Using
𝑤 𝑑 =^ 𝐹^ ∶^
1992 𝐽
. 75 𝑚 =^ 2656 N.
d. Using 𝑤 𝑑 =^ 𝐹^ ∶^
1992 𝐽
- 015 𝑚 =^ 132,800N.
9.2 Potential Energy
Practice
Questions
- What is the definition of energy?
- Name two types of potential energy.
- How is energy transferred from one object to another?
Answers
- 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
- 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?
- 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?
- 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
- Using 𝑃𝐸 = 𝑚𝑔ℎ ∶ 90.0𝑘𝑔 ∗ 9. 𝑚 𝑠^2 ∗ 9.47𝑚 ∶^ 8350 J.
- a. Using 𝑃𝐸 = 𝑚𝑔ℎ ∶ 50.0𝑘𝑔 ∗ 9. 𝑚 𝑠^2 ∗ 400𝑚 =^ 196,000 J. b. Using 𝑃𝐸 = 𝑚𝑔ℎ ∶ 50.0𝑘𝑔 ∗ 9. 𝑚 𝑠^2 ∗ 100𝑚 = 49,000𝐽^ Subtract initial PE from final PE : 196,000 − 49,000 = 147000 J.
- a. Using 𝑃𝐸 = (𝑚𝑔)ℎ ∶ 645𝑁 ∗ 4.55𝑚 = 2930 J. b. Starts from 0: 2930 J.
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
a. Using 𝑃𝐸 = 𝑚𝑔ℎ ∶ 15𝑘𝑔 ∗ 9. 𝑚 𝑠^2 ∗ 8𝑚 ∶^ 1180 J b. First solve for time to fall using: 𝑡 = √ 2𝑑 𝑔 ∶ √
2(8𝑚) 9.8𝑠𝑚 2 = 1.3𝑠. Then solve for velocity using: 𝑣 = 𝑔𝑡 ∶ 9.
𝑚 𝑠^2 ∗ 1.3𝑠 = 12.5 m/s
a. Solve for KE using 𝐾𝐸 = 1 2 𝑚𝑣
2 (85.0𝑘𝑔) (9.^
𝑚 𝑠 )
2 = 3442.5𝐽. Then solve for height using 3442.5J as PE with: 𝑃𝐸 𝑚𝑔 = ℎ ∶^
3442.5𝐽 85𝑘𝑔∗9.8 𝑚/𝑠^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.
- First solve for time to fall using: 𝑡 = √ 2𝑑 𝑔 ∶ √
2(9𝑚) 9.8𝑠𝑚 2 = 1.35𝑠. Then solve for velocity using: 𝑣 = 𝑔𝑡 ∶ 9. 𝑚 𝑠^2 ∗ 1.35𝑠 =13.23 m/s^ The performer’s velocity is 13.2 m/s at the bottom of the swing.
a. First solve for time to fall using: 𝑡 = √ 2𝑑 𝑔 ∶ √
2(45𝑚) 9.8𝑚𝑠 2 = 3.03𝑠. Then solve for velocity using: 𝑣 = 𝑔𝑡 ∶ 9. 𝑚 𝑠^2 ∗ 3.03𝑠 = 29.7 m/s
b. Speed at the top of hill#2 is equal to speed at the bottom – acceleration due to g over 40m. Acceleration over 40m = 𝑔 ∗ √ 2𝑑 𝑔 = 9.^
𝑚 𝑠^2 ∗ √
2∗40𝑚 9.8𝑠𝑚 2
𝑚 𝑠. 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
- Solve for final KE using 1 2 𝑚𝑣
2 (2𝑘𝑔) (^
𝑚 𝑠 )
2
(2𝑘𝑔) (9. 𝑚 𝑠^2 ) (20.0𝑚) = 792𝐽. Then solve for final velocity using^ 𝑣𝑓^ = √
2(792𝐽) 2.0𝑘𝑔 = √792 =^ 28.1 m/s.
- Using ℎ = 𝑃𝐸 𝑚𝑔 =^
500 2∗9.8 =^ 25.51 m.
9.4 Elastic and Inelastic Collisions
Practice
Questions
- Explain what happened in the first demonstration on elastic collisions.
- Explain what happened in the second demonstration on inelastic collisions.
- Assuming the first carts started at the same speed in both demonstrations, explain why the inelastic collision ended slower than the elastic collision.
Answers
- The momentum from the first cart was completely transferred into the second when the collision occurred and kinetic energy was unchanged.
- The momentum from the first cart was shared between the two carts when the collision occurred and kinetic energy decreased
- 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
- 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?
- Identify the following collisions as most likely elastic or most likely inelastic. a. A ball of modeling clay dropped on the floor.
