Energy Conservation and Transformations Energy Conservation and Transformations 1 / 10 A 10 kg object is dropped from a height of 20 m. What is the velocity just before it hits the ground? (g = 9.8 m/s²) 14 m/s 19.8 m/s 25 m/s 40 m/s Using energy conservation, v = √(2gh). v = √(2 × 9.8 × 20) = 19.8 m/s. Using energy conservation, v = √(2gh). v = √(2 × 9.8 × 20) = 19.8 m/s. 2 / 10 In a perfectly elastic collision, what is conserved? Momentum only Kinetic energy only Both momentum and kinetic energy Neither momentum nor kinetic energy 3 / 10 A system performs 500 J of work while releasing 300 J of heat to the surroundings. What is the change in internal energy? +200 J -200 J +800 J -800 J From the first law of thermodynamics, ΔU = Q – W = -300 – 500 = -200 J. From the first law of thermodynamics, ΔU = Q – W = -300 – 500 = -200 J. 4 / 10 A 50 kg skier slides down a frictionless slope of height 10 m. What is the skier’s velocity at the bottom? 14 m/s 20 m/s 22 m/s 10 m/s v = √(2gh) = √(2 × 9.8 × 10) = 14 m/s. v = √(2gh) = √(2 × 9.8 × 10) = 14 m/s. 5 / 10 A Carnot engine operates between temperatures of 600 K and 300 K. What is its efficiency? 25% 50% 75% 100% Efficiency = 1 – (Tc/Th) = 1 – (300/600) = 0.5 (50%). Efficiency = 1 – (Tc/Th) = 1 – (300/600) = 0.5 (50%). 6 / 10 A 5 kg block slides down a rough incline, losing 200 J of energy to friction. If the total mechanical energy at the top was 1000 J, what is its mechanical energy at the bottom? 1000 J 800 J 500 J 1200 J Mechanical energy at the bottom = Total energy – Energy lost to friction = 1000 – 200 = 800 J. Mechanical energy at the bottom = Total energy – Energy lost to friction = 1000 – 200 = 800 J. 7 / 10 A bullet of mass 0.05 kg traveling at 500 m/s embeds into a block of mass 5 kg at rest. What is their common velocity after the collision? 5 m/s 50 m/s 10 m/s 15 m/s Using momentum conservation: (0.05 × 500) = (5.05 × v) → v = 10 m/s. Using momentum conservation: (0.05 × 500) = (5.05 × v) → v = 10 m/s. 8 / 10 A spring is compressed by 0.2 m, storing 40 J of potential energy. What is the spring constant? 100 N/m 200 N/m 400 N/m 800 N/m PE = 1/2kx² → k = 2PE/x² = 2 × 40 / 0.2² = 400 N/m. PE = 1/2kx² → k = 2PE/x² = 2 × 40 / 0.2² = 400 N/m. 9 / 10 Which of the following best describes the transformation of energy in a hydroelectric power plant? Electrical to potential Kinetic to electrical Potential to kinetic to electrical Chemical to electrical 10 / 10 An athlete burns 500 calories (1 calorie = 4.184 J) while running. If this energy is used to lift a 50 kg mass, how high can it be lifted? 50 m 100 m 200 m 400 m Energy = mgh, so h = Energy/(mg) = 500 × 4.184 / (50 × 9.8) = 100 m. Energy = mgh, so h = Energy/(mg) = 500 × 4.184 / (50 × 9.8) = 100 m. Your score is By WordPress Quiz plugin
