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HW03_Sols_Giancoli

# HW03_Sols_Giancoli - Problem Set 3 Solutions(Giancoli 4th...

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Problem Set 3 Solutions (Giancoli 4 th ed.) 19. ( a ) Consider the free-body diagram for the crate on the surface. There is no motion in the y direction and thus no acceleration in the y direction. Write Newton’s second law for both directions, and find the acceleration. ( ) N N fr N cos 0 cos sin sin sin cos sin cos y x k k k F F mg F mg F mg F ma ma mg F mg mg a g ! ! ! ! μ ! μ ! ! μ ! = " = # = = + = = + = + = + \$ \$ Now use Eq. 2-12c, with an initial velocity of 3.0m s ! and a final velocity of 0 to find the distance the crate travels up the plane. ( ) ( ) ( ) ( ) 2 2 0 0 2 2 0 0 2 2 3.0m s 0.796m 2 2 9.80m s sin 25.0 0.17cos25.0 v v a x x v x x a ! = ! " ! ! ! ! = = = ! ° + ° The crate travels 0.80m up the plane. ( b ) We use the acceleration found above with the initial velocity in Eq. 2-12a to find the time for the crate to travel up the plane. ( ) ( ) ( ) 2 0 0 up 3.0m s 0.5308s 9.80m s sin 25.0 0.17cos25.0 up v v v at t a ! = = ° + ° + " = ! = ! The total time is NOT just twice the time to travel up the plane, because the acceleration of the block is different for the two parts of the motion. The second free-body diagram applies to the block sliding down the plane. A similar analysis will give the acceleration, and then Eq. 2-12b with an initial velocity of 0 is used to find the time to move down the plane. ( ) N N fr N cos 0 cos sin sin sin cos sin cos y x k k k F F mg F mg F mg F ma ma mg F mg mg a g ! ! ! ! μ ! μ ! ! μ ! = " = # = = " = = " = " = " \$ \$ ( ) ( ) ( ) ( ) 2 1 0 0 2 0 down 2 down up down 2 2 0.796m 0.7778s 9.80m s sin 25.0 0.17cos25.0 0.5308s 0.7778s 1.3s x x v t at x x t a t t t ! = + " ! = = = ° ! ° = + = + = m g ! N F ! fr F ! ! ! x y N F ! x y m g ! fr F ! ! !

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29. We assume that the child starts from rest at the top of the slide, and then slides a distance 0 x x ! along the slide. A force diagram is shown for the child on the slide. First, ignore the frictional force and so consider the no-friction case. All of the motion is in the x direction, so we will only consider Newton’s second law for the x direction. sin sin x F mg ma a g ! ! = = " = # Use Eq. 2-12c to calculate the speed at the bottom of the slide. ( ) ( ) ( ) ( ) 2 2 2 0 0 0 0 0 No friction 2 2 2 sin v v a x x v v a x x g x x ! " = " # = + " = " Now include kinetic friction. We must consider Newton’s second law in both the x and y directions now. The net force in the y direction must be 0 since there is no acceleration in the y direction.
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