This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY2048 Fall 2009 Homework Set 5 Solutions University of Florida Page 1 of 6 Department of Physics PHY 2048 Homework Set 5 Solutions (Chapter 6) Problem 65*: Here we sum all the vector forces on the crate and set the net force equal to the mass times the acceleration a m F F N i i net r r r = = = 1 (eq. 1) . Looking at the x component of (eq. 1) yields Ma F s = , where M is the mass of the crate and where I set a x = a ( i.e. deceleration). Looking at the y component of (eq. 1) yields = Mg F N and hence Mg F N = since a y = 0. We know that (F s ) max = s F N and hence 2 2 max max / 45 . 2 ) / 8 . 9 )( 25 . ( ) ( s m s m g M Mg M F M F a s s N s s = = = = = = . Using the kinematic equations of motion we have t a v t v x x max ) ( ) ( = 2 max 2 1 ) ( ) ( t a t v t x x = Where I let x = 0 at t = 0 and where s m km m s hr hr km v x / 33 . 13 10 600 , 3 1 / 48 ) ( 3 = = . Let t s be the time the train stops. Hence, s x s x t a v t v max ) ( ) ( = = and thus max ) ( a v t x s = . Let d be the distance travelled in the time t s . Hence, m s m s m a v t a t v t x d x s s x s 3 . 36 ) / 45 . 2 ( 2 ) / 33 . 13 ( 2 ) ( ) ( ) ( 2 2 max 2 2 max 2 1 = = = = = Problem 614*: Here we sum all the vector forces on the block and set the net force equal to the mass times the acceleration a m F F N i i net r r r = = = 1 (eq. 1) . First lets calculate the minimum static coeffiecient of friction such that the block does not move (i.e. a x = 0). Looking at the xcomponent of (eq. 1) yields cos = = x s Ma F F , where M is the mass of the block. Looking at the ycomponent of (eq. 1) yields sin = + Mg F F N since a y = 0. We know that F s = s F N and hence ) sin ( cos = F Mg F s (eq. 1) and solving for s gives F xaxis yaxis F N F g F k F s xaxis yaxis F N F g crate V PHY2048 Fall 2009 Homework Set 5 Solutions University of Florida Page 2 of 6 Department of Physics 5668 . ) 20 sin( 2 ) 20 cos( sin 5 . cos 5 . sin cos ) ( min = = = = o o Mg Mg Mg F Mg F s . (a) we see that s = 0.6 > 0.5668 and hence a x = 0....
View
Full
Document
This note was uploaded on 05/17/2011 for the course PHY 2048 taught by Professor Field during the Fall '08 term at University of Florida.
 Fall '08
 Field
 Physics, Acceleration, Force, Mass, Work

Click to edit the document details