Name Answer Key Date Unit 5 Review: Work, Energy & Power Academic Physics Part I: Conceptual 1. Define: a. Work: The product of a force applied over a distance b. Total (Mechanical) Energy: The Sum of Kinetic plus Potential energy c. Kinetic Energy: Energy of motion. d. Gravitational Potential Energy: Energy of position (height) e. Power: Rate at which energy is transferred/work is done 2. Circle all of the following situations where the specified force work on the specified object. I A student lifting a backpack with an upward applied force, II. The upward normal force on a hockey puck as it slides across the ice. The tension force acting on a pendulum as it swings back and forth IV. A student exerts an applied force to slide a cart across the floor. 3. A skateboarder starts at rest at point P. Describe the energy and motion of the skateboarder below at each point marked. Disregard the effects of friction. all P no k C B all K no P Some K Some P does not reach this Point all P ho k 4. What (in terms of energy conservation) would cause the skateboarder in the example above to slow down? The skateboarder slows down when the potential energy increases (this occurs when he is going uphill) The total Energy remains constant, so an increase in P causes a decrease in V. decrease in K, and therefore a m P (top) 5. A 20kg child starts from rest at the top of a slide with a potential energy of 500J What will be the velocity of the child at the bottom of the slide which is at ground level? P = mgh = E P = 0 Top: 05 + 5005 = 5005 Bottom: K = 500J on the ground. 500 J= 1/2 * (20kg) v^ 2 Rightarrow boxed v=7.07 M 5 6. In terms of energy, what would make choosing a small hybrid car a a ? Small hybrid, because it would require less work to change the kinetic energy. (thus accelerating the car) ← depends on mass W=K f -K i Rightarrow W=( 1/2 * m * v_{f} ^ 2 - 1 2 m v ; ^ 2 ) 7. Complete the table for the three different heights as the ball goes down the ramp. V_{0} = 0m / s 6.28 m K P mgh 9.53 kg mv² (9.55)(9.8) (14.14) 1321 J E K + P K+P= 13215 At rest E-K= 734J (9.53X9.8X6.28) 14.14 m = 587J 13215 E-K= 13215 At lowest Point 05 1321JV Part II Problems: Solve the following using methods demonstrated in class, SHOW ALL WORK. USE THE ENERGY EQUATIONS 8. Two 65kg students and a 12kg toboggan slide down a snow packed (frictionless) 12.4m high hill. If the sled starts from rest, what is its speed at the bottom of the hill? m = 65tan 65 * tan 12kg = 142kg h = 12.4m 12.4m TOP OJ BOTTOM 17,2565 mgh = 17,256 J 0 17,256J 17,256J k = 1/2 * m * v ^ 2 7, 256J = 1/2 * (142Vg) * v ^ 2 K + P = E 5ooJ + OJ = 500J V = 15.6 * m/5 mu = 1/2 * m * v ^ 2