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Unformatted text preview: 6.3: J 300 ) m . 12 )( N . 25 ( = . 6.18: As the example explains, the boats have the same kinetic energy K at the finish line, so 2 2 ) 2 / 1 ( ) 2 / 1 ( B B A A v m v m = , or, with 2 2 2 , 2 B A A B v v m m = = . a) Solving for the ratio of the speeds, 2 / = B A v v . b) The boats are said to start from rest, so the elapsed time is the distance divided by the average speed. The ratio of the average speeds is the same as the ratio of the final speeds, so the ratio of the elapsed times is 2 / / = = B A A B v v t t . 6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1) gives J. 1176 ) m . 25 )( s / m 80 . 9 )( kg 80 . 4 ( 2 = = = = mgs Fs W b) Since the melon is released from rest, 1 = K , and Eq. (6.6) gives J. 1176 2 = = = W K K 6.27: a) The friction force is mg k μ , which is directed against the car’s motion, so the net work done is mgs k μ- . The change in kinetic energy is 2 1 ) 2 / 1 ( mv K K- =- = ∆ , and so g v s k 2 2 / μ = . b) From the result of part (a), the stopping distance is proportional to the square of the initial speed, and so for an initial speed of 60 km/h, m 3 . 51 ) . 80 / . 60 )( m 2 . 91 ( 2 = = s . (This method avoids the intermediate calculation of k μ , which in this case is about 0.279.) 6.34: a) The average force is N 400 ) m 200 . /( ) J . 80 ( = , and the force needed to hold the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance quadruples the work so an extra 240 J of work must be done. The maximum force is quadrupled, 1600 N....
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