Lab 17: Energy and Levers ; Pendulum ; Mass and Spring - Force and distance in a lever - Behavior of a pendulum - Oscillations 19) A long crowbar is used to pry open an old window that is stuck shut. A force of 250 N was used to pull the crowbar a distance of 0.5 meter, after which the window was 2.5 cm open. What force did the crowbar exert on the window? Energy is conserved in a lever, so the energy = force distance generated on one end must equal the energy = force distance on the other end: energy = force distance = 250 N 0.5 m = 125 Joules You must convert cm to m for the second part: 2.5 cm = 0.025 m force = energy/distance = 125 Joules / 0.025 m = 5000 N (This is how a crowbar amplifies force when something is pried open.) 20) The following diagram shows a pendulum with an energy of 10 joules in various positions of its swing. a) Estimate the potential energy (PE) and kinetic energy (KE) in each position:
b) Given your answer to part a), is there a way to estimate the pendulumʼs speed in each position? Explain. Yes - if you know kinetic energy, you can find velocity. Just use the equation derived in problem 15)d) from kinetic energy = ½mass velocity 2 : velocity = (2 kinetic energy / mass)
Lab 18: A last look at force, motion and energy ; Air Ball - Projectiles - Kinetic and potential energy 21) A human cannonball (actually a spring launching a person inside of a fake cannon) launches a person through the air. a) When the person is standing in the cannon, well before the launch, what forces is he experiencing? Are these forces balanced? The person is experiencing a force of gravity (weight), and the bottom of the cannon (the spring) pushing back up. These forces are equal and opposite, and therefore balanced. b) The cannon launches the person into the air. While he is moving upward through the air, which direction is he accelerating? What forces does he experience? Are the forces acting on him balanced? Even though the person is moving upward (and sideways), the direction of acceleration is always down - as this is the direction toward which his speed is changing. The forces acting on him are no longer balanced, since there is now an acceleration - only gravity acts on the person. c) When the human cannonball begins moving downward, which direction is he accelerating? What forces does he experience (before landing)? Are the forces acting on him balanced? When he moves downward, he is still accelerating downward, just as he was before. The forces are still unbalanced; only gravity acts on him.
d) Referring back to Problem 10, when an astronaut is in orbit, which direction is she accelerating? What forces does she experience? Are the forces acting on her balanced? When an astronaut is in orbit, she is accelerating toward the earth ( downward ), just like the human cannonball. The force of gravity is acting on her, and, just like the human cannonball, these forces are not balanced, since she is accelerating. This is an important point. An accelerating falling body experiences unbalanced forces - there is no force pressing upwards on oneʼs body to make them feel weight, so there is a sensation of weightlessness, for both a human cannonball or an astronaut. Many people think that a astronauts feel weightless because they are too far from the earth to experience gravity - this is NOT true. Astronauts experience weightlessness because they are being accelerated toward the earth; astronauts actually NEED gravity if they are to remain in a stable orbit, as shown in Problem 10. (If an astronaut somehow stopped moving with a velocity perpendicular to her circular orbit, she would fall straight down like a meteor! Fortunately, the Law of Inertia prevents this...) 22) If you jumped out of a window, how far would you fall in 1 second? You observed and calculated in Lab 15 that gravity accelerates all falling objects (in absence of air resistance) at about 10 m/sec 2. acceleration = change in speed / time change in speed = acceleration time = 10 m/sec 2 1 sec = 10 m/sec The initial speed was zero, so the change give a final speed of 10 m/sec. To find the total distance, we need average speed, not final speed: average speed = ½(initial speed + final speed) = 0.5 (0 + 10 m/sec) = 5 m/sec average speed = total distance / total time total distance = average speed total time = 5 m/sec 1 sec = 5 meters