Chapter 10 - 10.1. Model: We will use the particle model...

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10.1. Model: We will use the particle model for the bullet (B) and the running student (S). Visualize: Solve: For the bullet, 22 B B B 11 (0.010 kg)(500 m/s) 1250 J K m v ! ! ! For the running student, S S S (75 kg)(5.5 m/s) 206 J K m v ! ! ! Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its speed is more than 90 times larger.
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10.2. Model: Model the hiker as a particle. Visualize: The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in Death Valley is 0 8.5 m. y !" Solve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is gf gi f i f i 26 () (65 kg)(9.8 m/s )[4420 m ( 85 m)] 2.9 10 J U U U mgy mgy mg y y # ! " ! " ! " ! " " ! $ Assess: Note that U # is independent of the origin of the coordinate system.
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10.3. Model: Model the compact car (C) and the truck (T) as particles. Visualize: Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal, 22 T C T C C T T C T C 1 1 20,000 kg (25 km/hr) 112 km/hr 2 2 1000 kg m K K m v m v v v m ! " ! " ! ! ! Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.
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10.4. Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Visualize: Solve: (a) The kinetic energy of the car is 2 2 5 C C C 11 (1500 kg)(30 m/s) 6.75 10 J 22 K m v ! ! ! " The car’s kinetic energy is 5 6.8 10 J. " (b) Let us relabel C K as f K and place our coordinate system at f 0 y ! m so that the car’s potential energy gf U is zero, its velocity is f , v and its kinetic energy is f . K At position i , y ii 0 m/s or 0 J, vK !! and the only energy the car has is gi i . U mgy ! Since the sum g K U # is unchanged by motion, f gf i gi . K U K U # ! # This means f f i i f i i 5 fi i 2 0 ( ) (6.75 10 J 0 J) 46 m (1500 kg)(9.8 m/s ) K mgy K mgy K K mgy KK y mg # ! # $ # ! # % " % $ ! ! ! (c) From part (b), &’ i () 2 mv mv vv y mg mg g % % % ! ! ! Free fall does not depend upon the mass.
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10.5. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls. Visualize: The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c). Solve: The quantity g K U ! is the same during free fall: f gf i gi . K U K U ! " ! We have (a) #$ 22 1 1 0 0 2 2 2 2 2 1 0 1 11 2 [(10 m/s) (0 m/s) ]/(2 9.8 m/s ) 5.10 m mv mgy mv mgy y v v g ! " ! % " & " & " 5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground. (b) 2 2 0 0 mv mgy mv mgy ! " ! Since 20 0, yy "" we get for the magnitudes 10 m/s. vv (c) 2 2 2 2 2 2 3 3 0 0 3 3 0 0 3 0 0 3 2 2 2 ( ) mv mgy mv mgy v gy v gy v v g y y ! " ! % ! " ! % " ! & 2 2 2 2 2 3 (10 m/s) 2(9.8 m/s )[0 m ( 20 m)] 492 m /s v % " ! & & " This means the magnitude of 3 v is equal to 22 m/s.
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Chapter 10 - 10.1. Model: We will use the particle model...

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