chapter 6 answers - 6.3(25.0 N(12.0 m 300 J 6.18 As the...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
6.3: J 300 ) m 0 . 12 )( N 0 . 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 0 . 25 )( s / m 80 . 9 )( kg 80 . 4 ( 2 = = = = mgs Fs W b) Since the melon is released from rest, 0 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 0 1 ) 2 / 1 ( mv K K - = - = , and so g v s k 2 0 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 ) 0 . 80 / 0 . 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 . 0 /( ) J 0 . 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.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern