2014_SP08_Sol07

2014_SP08_Sol07 - 10.5. Model: Model the car (C) as a...

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10.5. 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 1 1 (1500 kg)(30 m/s) 6.75 10 J 2 2 K m v = = = × (b) Let us relabel K C as K f and place our coordinate system at f 0 y = m so that the car’s potential energy U gf is zero, its velocity is v f , and its kinetic energy is K f . At position y i , i i 0 m/s or 0 J, v K = = and the only energy the car has is gi i . U mgy = Since the sum K + U g is unchanged by motion, f gf i gi K U K U + = + . This means f f i i f i i 5 f i i 2 0 ( ) (6.75 10 J 0 J) 45.9 m (1500 kg)(9.8 m/s ) K mgy K mgy K K mgy K K y mg + = + + = + - × - = = = (c) From part (b), ( 29 2 2 2 2 f i f i f i i 1 1 ( ) 2 2 2 mv mv v v K K y mg mg g - - - = = = Free fall does not depend upon the mass. 10.11. Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change as the pendulum swings from one side to the other.
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Visualize: The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: (a) The quantity K + U g is the same at the lowest point of the trajectory as it was at the highest point. Thus, 1 g1 0 g0 means K U K U + = + 2 2 2 2 1 1 0 0 1 1 0 0 2 2 1 0 1 0 1 1 2 2 2 2 2 (0 m) (0 m/s) 2 2 mv mgy mv mgy v gy v gy v g gy v gy + = + + = + + = + = From the pictorial representation, we find that 0 cos30 . y L L = - ° Thus, 2 1 2 (1 cos30 ) 2(9.8 m/s )(0.75 m)(1 cos30 )
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2014_SP08_Sol07 - 10.5. Model: Model the car (C) as a...

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