CHAPTER 4 - 2 - 9.

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

View Full Document Right Arrow Icon
F T m g y 9. The required average acceleration can be found from the one-dimensional motion: v 2  =  v 0 2  + 2 a ( x  –  x 0 ); 0 = [(100 km/h)/(3.6 ks/h)] 2  + 2 a (150 m – 0), which gives  a  = – 2.57 m/s 2 . We apply Newton’s second law to find the required force ? ma ;   F  = (3.6 × 10 5  kg)(– 2.57 m/s 2 ) =         – 9.3 × 10 5  N . The weight of the train is mg  = (3.6 × 10 5  kg)(9.80 m/s 2 ) = 3.5 × 10 6  N,  so Superman must apply a force that is       25%       of the weight of the train. 10. The acceleration of the first box is a 1  =  F / m 1   , so the speed after a time  t  is  v 1  =  v 0  +  a 1 t  =  a 1 t . For the second box we have a 2  =  F / m 2    =  F /2 m 1  =   ! a 1   ,  so the speed after a time  t  is  v 2  =  v 0  +  a 2 t  =  ! a 1 t ,  or         v 2  =  ! v 1 . 11.
Background image of page 1

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

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

This note was uploaded on 03/08/2010 for the course PHYSICS 7A/7B taught by Professor All during the Fall '08 term at Berkeley.

Page1 / 2

CHAPTER 4 - 2 - 9.

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online