Study Translational Kinetic Energy in AP Physics 1 with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: What does m m m represent in the translational kinetic energy formula? Answer: Mass of the object. Mass is the scalar quantity representing the amount of matter in the object.
Flashcard 2: Determine the kinetic energy if mass is 10 kg and velocity is 0. Answer: K E = 0 J KE = 0 \text{ J} K E = 0 J . Any object at rest has zero kinetic energy.
Flashcard 3: What is the kinetic energy formula rearranged for velocity? Answer: v = 2 K E m v = \sqrt{\frac{2KE}{m}} v = m 2 K E . Solve K E = 1 2 m v 2 KE = \frac{1}{2}mv^2 K E = 2 1 m v 2 for v v v by isolating the velocity term.
Flashcard 4: If kinetic energy is 200 J and velocity is 4 m/s, find the mass. Answer: m = 25 kg m = 25 \text{ kg} m = 25 kg . 200 = 1 2 m ( 4 2 ) 200 = \frac{1}{2}m(4^2) 200 = 2 1 m ( 4 2 ) , so m = 2 ( 200 ) 16 = 25 m = \frac{2(200)}{16} = 25 m = 16 2 ( 200 ) = 25 kg.
Flashcard 5: A car increases speed from 10 m/s to 20 m/s. By what factor does its kinetic energy increase? Answer: Increases by a factor of 4. Velocity doubles, so K E KE K E increases by ( 2 ) 2 = 4 (2)^2 = 4 ( 2 ) 2 = 4 times.
Flashcard 6: If velocity is doubled, how does kinetic energy change? Answer: Increases by a factor of 4. Since v 2 v^2 v 2 appears in the formula, doubling v v v increases K E KE K E by 2 2 = 4 2^2 = 4 2 2 = 4 .
Flashcard 7: Find kinetic energy: mass = 0.5 kg, velocity = 10 m/s. Answer: K E = 25 J KE = 25 \text{ J} K E = 25 J . K E = 1 2 ( 0.5 ) ( 10 2 ) = 1 2 ( 0.5 ) ( 100 ) = 25 KE = \frac{1}{2}(0.5)(10^2) = \frac{1}{2}(0.5)(100) = 25 K E = 2 1 ( 0.5 ) ( 1 0 2 ) = 2 1 ( 0.5 ) ( 100 ) = 25 J.
Flashcard 8: State the relationship between kinetic energy and velocity. Answer: Proportional to the square of velocity. The v 2 v^2 v 2 term creates a quadratic relationship with velocity.
Flashcard 9: What are the SI units of translational kinetic energy? Answer: Joules (J). Energy units derived from k g ⋅ m 2 / s 2 kg \cdot m^2/s^2 k g ⋅ m 2 / s 2 in the SI system.
Flashcard 10: Convert 1 Joule into base SI units. Answer: 1 J = 1 kg × m 2 / s 2 1 \text{ J} = 1 \text{ kg} \times \text{m}^2/\text{s}^2 1 J = 1 kg × m 2 / s 2 . Breaking down the joule into fundamental SI base units.
Flashcard 11: Calculate the mass of an object with 50 J of kinetic energy moving at 5 m/s. Answer: m = 4 kg m = 4 \text{ kg} m = 4 kg . 50 = 1 2 m ( 5 2 ) 50 = \frac{1}{2}m(5^2) 50 = 2 1 m ( 5 2 ) , so m = 2 ( 50 ) 25 = 4 m = \frac{2(50)}{25} = 4 m = 25 2 ( 50 ) = 4 kg.
Flashcard 12: If mass is tripled and velocity is constant, how does K E KE K E change? Answer: Triples. Kinetic energy scales linearly with mass when velocity is constant.
Flashcard 13: A 1 kg object moves at 3 m/s. Find its kinetic energy. Answer: K E = 4.5 J KE = 4.5 \text{ J} K E = 4.5 J . K E = 1 2 ( 1 ) ( 3 2 ) = 1 2 ( 1 ) ( 9 ) = 4.5 KE = \frac{1}{2}(1)(3^2) = \frac{1}{2}(1)(9) = 4.5 K E = 2 1 ( 1 ) ( 3 2 ) = 2 1 ( 1 ) ( 9 ) = 4.5 J.
Flashcard 14: What is the kinetic energy formula rearranged for velocity? Answer: v = sqrt ( 2 K E m ) v = \text{sqrt}(\frac{2KE}{m}) v = sqrt ( m 2 K E ) . Solve K E = 1 2 m v 2 KE = \frac{1}{2}mv^2 K E = 2 1 m v 2 for v v v by isolating the velocity term.
Flashcard 15: Calculate velocity: kinetic energy = 18 J, mass = 2 kg. Answer: v = 3 m/s v = 3 \text{ m/s} v = 3 m/s . 18 = 1 2 ( 2 ) v 2 18 = \frac{1}{2}(2)v^2 18 = 2 1 ( 2 ) v 2 , so v = 36 2 = 3 v = \sqrt{\frac{36}{2}} = 3 v = 2 36 = 3 m/s.
Flashcard 16: A car increases speed from 10 m/s to 20 m/s. By what factor does its kinetic energy increase? Answer: Increases by a factor of 4. Velocity doubles, so K E KE K E increases by ( 2 ) 2 = 4 (2)^2 = 4 ( 2 ) 2 = 4 times.
Flashcard 17: Convert 100 J to kilojoules (kJ). Answer: 0.1 kJ 0.1 \text{ kJ} 0.1 kJ . Divide by 1000 to convert joules to kilojoules.
Flashcard 18: A 0.5 kg object has 8 J of kinetic energy. What is its velocity? Answer: v = 4 m/s v = 4 \text{ m/s} v = 4 m/s . 8 = 1 2 ( 0.5 ) v 2 8 = \frac{1}{2}(0.5)v^2 8 = 2 1 ( 0.5 ) v 2 , so v = 16 0.5 = 4 v = \sqrt{\frac{16}{0.5}} = 4 v = 0.5 16 = 4 m/s.
Flashcard 19: What factor affects translational kinetic energy the most? Answer: Velocity, as v v v is squared. The squared term makes velocity changes more impactful than mass changes.
Flashcard 20: What is the formula to find mass from kinetic energy and velocity? Answer: m = 2 K E v 2 m = \frac{2KE}{v^2} m = v 2 2 K E . Rearrange K E = 1 2 m v 2 KE = \frac{1}{2}mv^2 K E = 2 1 m v 2 to solve for mass.
Flashcard 21: Calculate the kinetic energy: mass = 4 kg, velocity = 5 m/s. Answer: K E = 50 J KE = 50 \text{ J} K E = 50 J . K E = 1 2 ( 4 ) ( 5 2 ) = 1 2 ( 4 ) ( 25 ) = 50 KE = \frac{1}{2}(4)(5^2) = \frac{1}{2}(4)(25) = 50 K E = 2 1 ( 4 ) ( 5 2 ) = 2 1 ( 4 ) ( 25 ) = 50 J.
Flashcard 22: What is the kinetic energy of a 3 kg object moving at 2 m/s? Answer: K E = 6 J KE = 6 \text{ J} K E = 6 J . K E = 1 2 ( 3 ) ( 2 2 ) = 1 2 ( 3 ) ( 4 ) = 6 KE = \frac{1}{2}(3)(2^2) = \frac{1}{2}(3)(4) = 6 K E = 2 1 ( 3 ) ( 2 2 ) = 2 1 ( 3 ) ( 4 ) = 6 J.
Flashcard 23: If mass is halved, how does kinetic energy change? Answer: Halved. Kinetic energy is directly proportional to mass.
Flashcard 24: What is the kinetic energy of a 5 kg object at rest? Answer: K E = 0 J KE = 0 \text{ J} K E = 0 J . Zero velocity means zero kinetic energy regardless of mass.
Flashcard 25: Calculate the kinetic energy: mass = 4 kg, velocity = 5 m/s. Answer: K E = 50 J KE = 50 \text{ J} K E = 50 J . K E = 1 2 ( 4 ) ( 5 2 ) = 1 2 ( 4 ) ( 25 ) = 50 KE = \frac{1}{2}(4)(5^2) = \frac{1}{2}(4)(25) = 50 K E = 2 1 ( 4 ) ( 5 2 ) = 2 1 ( 4 ) ( 25 ) = 50 J.
Flashcard 26: A 3 kg object has 27 J of kinetic energy. Find its velocity. Answer: v = 3 m/s v = 3 \text{ m/s} v = 3 m/s . 27 = 1 2 ( 3 ) v 2 27 = \frac{1}{2}(3)v^2 27 = 2 1 ( 3 ) v 2 , so v = 54 3 = 3 v = \sqrt{\frac{54}{3}} = 3 v = 3 54 = 3 m/s.
Flashcard 27: If kinetic energy is 200 J and velocity is 4 m/s, find the mass. Answer: m = 25 kg m = 25 \text{ kg} m = 25 kg . 200 = 1 2 m ( 4 2 ) 200 = \frac{1}{2}m(4^2) 200 = 2 1 m ( 4 2 ) , so m = 2 ( 200 ) 16 = 25 m = \frac{2(200)}{16} = 25 m = 16 2 ( 200 ) = 25 kg.
Flashcard 28: Determine the kinetic energy if mass is 10 kg and velocity is 0. Answer: K E = 0 J KE = 0 \text{ J} K E = 0 J . Any object at rest has zero kinetic energy.
Flashcard 29: Determine the kinetic energy change if velocity triples. Answer: Increases by a factor of 9. Tripling velocity increases K E KE K E by ( 3 ) 2 = 9 (3)^2 = 9 ( 3 ) 2 = 9 times.
Flashcard 30: If kinetic energy is 64 J and mass is 4 kg, what is the velocity? Answer: v = 4 m/s v = 4 \text{ m/s} v = 4 m/s . 64 = 1 2 ( 4 ) v 2 64 = \frac{1}{2}(4)v^2 64 = 2 1 ( 4 ) v 2 , so v = 128 4 = 4 v = \sqrt{\frac{128}{4}} = 4 v = 4 128 = 4 m/s.
Flashcard 31: Identify the SI unit for mass used in the kinetic energy formula. Answer: Kilogram (kg). Standard SI base unit for measuring the amount of matter.
Flashcard 32: Find kinetic energy: mass = 0.5 kg, velocity = 10 m/s. Answer: K E = 25 J KE = 25 \text{ J} K E = 25 J . K E = 1 2 ( 0.5 ) ( 10 2 ) = 1 2 ( 0.5 ) ( 100 ) = 25 KE = \frac{1}{2}(0.5)(10^2) = \frac{1}{2}(0.5)(100) = 25 K E = 2 1 ( 0.5 ) ( 1 0 2 ) = 2 1 ( 0.5 ) ( 100 ) = 25 J.
Flashcard 33: If velocity remains constant, how does kinetic energy change as mass increases? Answer: Increases linearly. Direct proportional relationship when velocity remains constant.
Flashcard 34: What factor affects translational kinetic energy the most? Answer: Velocity, as v v v is squared. The squared term makes velocity changes more impactful than mass changes.
Flashcard 35: A 10 kg object moves at 0 m/s. What is its kinetic energy? Answer: K E = 0 J KE = 0 \text{ J} K E = 0 J . Zero velocity means zero kinetic energy regardless of mass.
Flashcard 36: Convert 100 J to kilojoules (kJ). Answer: 0.1 kJ 0.1 \text{ kJ} 0.1 kJ . Divide by 1000 to convert joules to kilojoules.
Flashcard 37: What is the kinetic energy of a 3 kg object moving at 2 m/s? Answer: K E = 6 J KE = 6 \text{ J} K E = 6 J . K E = 1 2 ( 3 ) ( 2 2 ) = 1 2 ( 3 ) ( 4 ) = 6 KE = \frac{1}{2}(3)(2^2) = \frac{1}{2}(3)(4) = 6 K E = 2 1 ( 3 ) ( 2 2 ) = 2 1 ( 3 ) ( 4 ) = 6 J.
Flashcard 38: What are the SI units of translational kinetic energy? Answer: Joules (J). Energy units derived from k g ⋅ m 2 / s 2 kg \cdot m^2/s^2 k g ⋅ m 2 / s 2 in the SI system.
Flashcard 39: If velocity remains constant, how does kinetic energy change as mass increases? Answer: Increases linearly. Direct proportional relationship when velocity remains constant.
Flashcard 40: If kinetic energy is 64 J and mass is 4 kg, what is the velocity? Answer: v = 4 m/s v = 4 \text{ m/s} v = 4 m/s . 64 = 1 2 ( 4 ) v 2 64 = \frac{1}{2}(4)v^2 64 = 2 1 ( 4 ) v 2 , so v = 128 4 = 4 v = \sqrt{\frac{128}{4}} = 4 v = 4 128 = 4 m/s.
Flashcard 41: Identify the SI unit for velocity in the kinetic energy formula. Answer: Meters per second (m/s). SI derived unit combining distance per unit time.
Flashcard 42: A 10 kg object moves at 0 m/s. What is its kinetic energy? Answer: K E = 0 J KE = 0 \text{ J} K E = 0 J . Zero velocity means zero kinetic energy regardless of mass.
Flashcard 43: State the relationship between kinetic energy and mass. Answer: Directly proportional. Doubling mass doubles kinetic energy when velocity is constant.
Flashcard 44: Calculate the kinetic energy of a 2 kg object moving at 3 m/s. Answer: K E = 9 J KE = 9 \text{ J} K E = 9 J . K E = 1 2 ( 2 ) ( 3 2 ) = 1 2 ( 2 ) ( 9 ) = 9 KE = \frac{1}{2}(2)(3^2) = \frac{1}{2}(2)(9) = 9 K E = 2 1 ( 2 ) ( 3 2 ) = 2 1 ( 2 ) ( 9 ) = 9 J.
Flashcard 45: How is translational kinetic energy related to work done? Answer: Equal to work done to bring object to current speed. Work-energy theorem states work done equals change in kinetic energy.
Flashcard 46: What is the kinetic energy of a 2 kg object moving at 0 m/s? Answer: K E = 0 J KE = 0 \text{ J} K E = 0 J . Zero velocity results in zero kinetic energy.
Flashcard 47: Convert 1 Joule into base SI units. Answer: 1 J = 1 kg × m 2 / s 2 1 \text{ J} = 1 \text{ kg} \times \text{m}^2/\text{s}^2 1 J = 1 kg × m 2 / s 2 . Breaking down the joule into fundamental SI base units.
Flashcard 48: State the relationship between kinetic energy and velocity. Answer: Proportional to the square of velocity. The v 2 v^2 v 2 term creates a quadratic relationship with velocity.
Flashcard 49: A 0.5 kg object has 8 J of kinetic energy. What is its velocity? Answer: v = 4 m/s v = 4 \text{ m/s} v = 4 m/s . 8 = 1 2 ( 0.5 ) v 2 8 = \frac{1}{2}(0.5)v^2 8 = 2 1 ( 0.5 ) v 2 , so v = 16 0.5 = 4 v = \sqrt{\frac{16}{0.5}} = 4 v = 0.5 16 = 4 m/s.
Flashcard 50: What happens to kinetic energy if both mass and velocity are doubled? Answer: Increases by a factor of 8. Mass doubles (× 2 \times 2 × 2 ) and velocity squared doubles (× 4 \times 4 × 4 ): 2 × 4 = 8 2 \times 4 = 8 2 × 4 = 8 .
Flashcard 51: A 3 kg object has 27 J of kinetic energy. Find its velocity. Answer: v = 3 m/s v = 3 \text{ m/s} v = 3 m/s . 27 = 1 2 ( 3 ) v 2 27 = \frac{1}{2}(3)v^2 27 = 2 1 ( 3 ) v 2 , so v = 54 3 = 3 v = \sqrt{\frac{54}{3}} = 3 v = 3 54 = 3 m/s.
Flashcard 52: What does m m m represent in the translational kinetic energy formula? Answer: Mass of the object. Mass is the scalar quantity representing the amount of matter in the object.
Flashcard 53: What does v v v represent in the translational kinetic energy formula? Answer: Velocity of the object. Velocity is the vector quantity representing the object's speed and direction.
Flashcard 54: State the formula for translational kinetic energy. Answer: K E = 1 2 m v 2 KE = \frac{1}{2}mv^2 K E = 2 1 m v 2 . Fundamental formula where kinetic energy equals half the product of mass and velocity squared.
Flashcard 55: What happens to kinetic energy if both mass and velocity are doubled? Answer: Increases by a factor of 8. Mass doubles (× 2 \times 2 × 2 ) and velocity squared doubles (× 4 \times 4 × 4 ): 2 × 4 = 8 2 \times 4 = 8 2 × 4 = 8 .
Flashcard 56: Calculate velocity: kinetic energy = 18 J, mass = 2 kg. Answer: v = 3 m/s v = 3 \text{ m/s} v = 3 m/s . 18 = 1 2 ( 2 ) v 2 18 = \frac{1}{2}(2)v^2 18 = 2 1 ( 2 ) v 2 , so v = 36 2 = 3 v = \sqrt{\frac{36}{2}} = 3 v = 2 36 = 3 m/s.
Flashcard 57: How is translational kinetic energy related to work done? Answer: Equal to work done to bring object to current speed. Work-energy theorem states work done equals change in kinetic energy.
Flashcard 58: A 6 kg object moving at 2 m/s. Calculate its kinetic energy. Answer: K E = 12 J KE = 12 \text{ J} K E = 12 J . K E = 1 2 ( 6 ) ( 2 2 ) = 1 2 ( 6 ) ( 4 ) = 12 KE = \frac{1}{2}(6)(2^2) = \frac{1}{2}(6)(4) = 12 K E = 2 1 ( 6 ) ( 2 2 ) = 2 1 ( 6 ) ( 4 ) = 12 J.
Flashcard 59: If mass is tripled and velocity is constant, how does K E KE K E change? Answer: Triples. Kinetic energy scales linearly with mass when velocity is constant.
Flashcard 60: Calculate the mass of an object with 50 J of kinetic energy moving at 5 m/s. Answer: m = 4 kg m = 4 \text{ kg} m = 4 kg . 50 = 1 2 m ( 5 2 ) 50 = \frac{1}{2}m(5^2) 50 = 2 1 m ( 5 2 ) , so m = 2 ( 50 ) 25 = 4 m = \frac{2(50)}{25} = 4 m = 25 2 ( 50 ) = 4 kg.
Flashcard 61: Identify the SI unit for mass used in the kinetic energy formula. Answer: Kilogram (kg). Standard SI base unit for measuring the amount of matter.
Flashcard 62: Determine the kinetic energy change if velocity triples. Answer: Increases by a factor of 9. Tripling velocity increases K E KE K E by ( 3 ) 2 = 9 (3)^2 = 9 ( 3 ) 2 = 9 times.