{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

05_InstSolManual_PDF_Part17 - and m k2 5 m k1 1 v 02 v 01 2...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
5.36. Set Up: Use to find the acceleration that can be given to the car by the kinetic friction force. Then use a constant acceleration equation. Take in the direction the car is moving. Solve: (a) The free-body diagram for the car is shown in Figure 5.36. gives gives and Then and gives Figure 5.36 (b) 5.37. Set Up: The free-body diagram for the car is shown in Figure 5.37. Figure 5.37 Solve: Relate a to gives gives and Use a kinematic equation to relate a to the stopping distance d . (stops), gives Same means and and so Reflect: The stopping distance is proportional to the acceleration and the acceleration is proportional to the friction force, so the stopping distance is decreased by the same factor as is the coefficient of friction. 5.38. Set Up: The free-body diagram for the car is given in Figure 5.37. Solve: As in Problem 5.37, Same D means and
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: and m k2 5 m k1 1 v 02 v 01 2 2 5 m 1 3 V V 2 2 5 9 m . v 01 2 m k1 5 v 02 2 m k2 v 2 5 2 aD 5 2 m k gD . a 5 m k g . d w 5 8 D . m kw 5 1 8 m kd d d 5 D d w 5 1 m kd m kw 2 d d . m kd d d 5 m kw d w . a d d d 5 a w d w v v 2 5 1 2 ad . 2 a x 1 x 2 x 2 v x 2 5 v x 2 1 x 2 x 5 d . v x 5 a x 5 2 a , a 5 m k g . m k mg 5 ma 2 f k 5 2 ma . g F x 5 ma x n 5 mg . g F y 5 ma y m k . y x mg n f k v a v x 5 " 2 m k g 1 x 2 x 2 5 " 2 1 0.25 21 9.80 m / s 2 21 54.0 m 2 5 16.3 m / s 1 x 2 x 2 5 2 v x 2 2 a x 5 1 v x 2 2 m k g 5 1 29.1 m / s 2 2 2 1 0.80 21 9.80 m / s 2 2 5 54.0 m. y x mg n f k 5 m k n v v x 2 5 v x 2 1 2 a x 1 x 2 x 2 v x 5 a x 5 2m k g . 2m k mg 5 ma x 2m k n 5 ma x . g F x 5 ma x n 5 mg . g F y 5 ma y 1 x g F S 5 m a S Applications of Newton’s Laws 5-17...
View Full Document

{[ snackBarMessage ]}