Physics Solution Manual for 1100 and 2101

3 ms chapter 6 problems 307 b the calculation and

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Unformatted text preview: can be found from Equation 2.9 with v0 = 0 m/s, since the rock is released from rest. At a height h, the rock has fallen through a distance 2 (20 .0 m) – h , and according to Equation 2.9, v = 2ay = 2 a[(20.0 m) – h]. Therefore, the kinetic energy at any height h is given by KE = ma[(20.0 m)– h ]. The total energy at any height h is the sum of the kinetic energy and potential energy at the particular height. SOLUTION The calculations are performed below for h = 10.0 m . The table that follows also shows the results for h = 20.0 m and h = 0 m. 2 PE = mgh = (2.00 kg)(9.80 m/s )(10.0 m) = 196 J KE = ma[ (20.0 m) – h ] = (2.00 kg)(9.80 m/s 2 )(20.0 m – 10.0 m) = 196 J E = KE + PE = 196 J + 196 J = h (m) 20.0 10.0 0 KE (J) 0 196 392 PE (J) 392 196 0 392 J E (J) 392 392 392 Note that in each case, the value of E is the same, because mechanical energy is conserved. ______________________________________________________________________________ 38. REASONING a. and b. Since there is no friction, there are only two forces acting on each box, the gravitational force (its weight) and the normal force. The normal force, being perpendicular to the displacement of a box, does no work. The gravitational force does work. However, it is a conservative force, so the principle of conservation of mechanical energy applies to the motion of each box, and we can use it to determine the speed in each case. This principle reveals that the final speed depends only on the initial speed, and the initial and final heights. The final speed does not depend on the mass of the box or on the steepness of the slope. Thus, both boxes reach B with the same speed. c. The kinetic energy is given by the expression KE = 1 mv 2 (Equation 6.2). Both boxes 2 have the same speed v when they reach B. However, the heavier box has the greater mass. Therefore, the heavier box has the greater kinetic energy. 306 WORK AND ENERGY SOLUTION a. The conservation of mechanical energy states that 1 2 2 mvB + mghB = 1 mv 2 A 2 + mghA (6.9b) Solving for the speed vB at B gives vB = vA + 2 g ( hA − hB ) = 2 ( 0 m /s )2 + 2 ( 9.80 m / s 2 ) ( 4.5 m − 1.5 m ) = 7.7 m /s b. The speed of the heavier box is the same as that of the lighter box: 7.7 m /s . c. The ratio of the kinetic energies is ( KEB )heavier box ( KEB )lighter box ( = ( ) ) 1 mv 2 B 2 heavier box 2 1 mv B 2 lighter box = mheavier box mlighter box = 44 kg = 11 kg 4.0 ______________________________________________________________________________ 39. REASONING We can find the landing speed vf from the final kinetic energy KEf, since KEf = 1 mvf2 . Furthermore, we are ignoring air resistance, so the conservation of mechanical 2 energy applies. This principle states that the mechanical energy with which the pebble is initially launched is also the mechanical energy with which it finally strikes the ground. Mechanical energy is kinetic energy plus gravitational potential energy. With respect to potential energy we will use ground level as the zero level for measuring heights. The pebble initially has kinetic and potential energies. When it strikes the ground, however, the pebble has only kinetic energy, its potential energy having been converted into kinetic energy. It does not matter how the pebble is launched, because only the vertical height determines the gravitational potential energy. Thus, in each of the three parts of the problem the same amount of potential energy is converted into kinetic energy. As a result, the speed that we will calculate from the final kinetic energy will be the same in parts (a), (b), and (c). SOLUTION a. Applying the conservation of mechanical energy in the form of Equation 6.9b, we have 2 2 1 1 2 mvf + mgh = 2 mv0 + mgh0 14 244f 4 3 14 244 4 3 Final mechanical energy at ground level Initial mechanical energy at top of building Solving for the final speed gives 2 vf = v0 + 2 g ( h0 − hf ) = (14.0 m/s ) 2 + 2 ( 9.80 m/s 2 ) ( 31.0 m ) − ( 0 m ) = 28.3 m/s Chapter 6 Problems 307 b. The calculation and answer are the same as in part (a). c. The calculation and answer are the same as in part (a). 40. REASONING The distance h in the drawing in the text is the difference between the skateboarder’s final and initial heights (measured, for example, with respect to the ground), or h = hf − h0. The difference in the heights can be determined by using the conservation of mechanical energy. This conservation law is applicable because nonconservative forces are negligible, so the work done by them is zero (Wnc = 0 J). Thus, the skateboarder’s final total mechanical energy Ef is equal to his initial total mechanical energy E0: 1 mv 2 + mgh = 1 mv 2 + mgh f 244f 02440 2 2 144 3 144 3 Ef (6.9b) E0 Solving Equation 6.9b for hf − h0, we find that hf − h0 = 1 24 43 h 1 v2 20 − 1 vf2 2 g SOLUTION Using the fact that v0 = 5.4 m/s and vf = 0 m/s (since the skateboarder comes to a momentary rest), the distance h is h= 1 v2 20 1 ( 5.4 2 − 1 vf2 2 2 m/s ) − 1 ( 0 m/s ) 2 2 = = 1.5 m g 9.80 m/s 2 ______________________________________________________________________________ 41. REASONING Since air resistance is being...
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