Chapter 1 Conservation of Mass - Light and Matter

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of gravity doesn’t change much if you only move up or down a few meters. We also find that the gravitational energy is proportional to the mass of the object we’re testing. Writing y for the height, and g for the overall constant of proportionality, we have Ug = mgy . [gravitational energy; y=height; only accurate within a small range of heights] The number g, with units of joules per kilogram per meter, is called the gravitational field. It tells us the strength of gravity in a certain region of space. Near the surface of our planet, it has a value of about 9.8 J/kg · m, which is conveniently close to 10 J/kg · m for rough calculations. Velocity at the bottom of a drop example 6 ⊲ If the skater in figure g drops 3 meters from rest, what is his velocity at the bottom of the pool? ⊲ Starting from conservation of energy, we have so 0 = ∆E = ∆K + ∆U = K f − K i + U f − U i = 1 2 mv f 2 + mgy f − mgy i (because K i=0) = 1 2 mv f 2 + mg∆y , (∆y
elated to gravitational mass, but since these two masses are equal, we were able to use a single symbol, m, for them, and cancel them out. We can see from the equation v = √ −2g∆y that a falling object’s velocity isn’t constant. It increases as the object drops farther and farther. What about its acceleration? If we assume that air friction is negligible, the arguments in section 1.2 show that the acceleration can’t depend on the object’s mass, so there isn’t much else the acceleration can depend on besides g. In fact, the acceleration of a falling object equals −g (in a coordinate system where the positive y axis points up), as we can easily show using the chain rule: � � � � � � � � � � dv dv dK dU dy = dt dK dU dy dt � � 1 = (−1)(mg)(v) mv = −g , where I’ve calculated dv/dK as 1/(dK/dv), and dK/dU = −1 can be found by differentiating K+U = (constant) to give dK+dU = 0. 7 We can also check that the units of g, J/kg·m, are equivalent to the units of acceleration, J kg·m = kg·m2 /s2 kg·m = m s 2 and therefore the strength of the gravitational field near the earth’s surface can just as well be stated as 10 m/s 2 . Speed after a given time example 7 ⊲ An object falls from rest. How fast is it moving after two seconds? Assume that the amount of energy converted to heat by air friction is negligible. ⊲ Under the stated assumption, we have a = −g, which can be integrated to give v = −gt + constant. If we let t = 0 be the beginning of the fall, then the constant of integration is zero, so at t = 2 s we have v = −gt = −(10 m/s 2 ) × (2 s) = 20 m/s. The Vomit Comet example 8 ⊲ The U.S. Air Force has an airplane, affectionately known as the Vomit Comet, in which astronaut trainees can experience simulated weightlessness. Oversimplifying a little, imagine that the plane climbs up high, and then drops straight down like a rock. (It actually flies along a parabola.) Since the people are falling with the same acceleration as the plane, the sensation is just like what you’d experience if you went out of the earth’s gravitational 7 There is a mathematical loophole in this argument that would allow the object to hover for a while with zero velocity and zero acceleration. This point is discussed on page 898. , Section 2.1 Energy 83
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Chapter 1 Conservation of Mass - Light and Matter

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