166_PartUniversity Physics Solution

# 166_PartUniversity Physics Solution - 5-38 Chapter 5 2 2 v...

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5-38 Chapter 5 0 0, y = 22 00 2( ) yy y vv a =+ gives 2 0 2 0 0(7 4 m s ) 5.95 m s 2( ) 460 m y vv a −− == = , which is the vertical acceleration that must be provided by the PAPS. The time it takes to reach the ground is given by 0 2 4 m s ) 12.4 s 5.95 m s y t a = Using Newton&s Second Law for the vertical direction PAPSv Fm g m a += . This gives 2 PAPSv ( ) (150 kg)(5.95 ( 3.7)) m s 1450 N a m g m a g =−= + = = , which is the vertical component of the PAPS force. The vehicle must also be brought to a stop horizontally in 12.4 seconds; the acceleration needed to do this is 2 0 2 0 33 m s 2.66 m s 12.4 s y a t = and the force needed is 2 PAPSh (150 kg)(2.66 m s ) 400 N a = , since there are no other horizontal forces. EVALUATE: The acceleration produced by the PAPS must bring to zero both her horizontal and vertical components of velocity. 5.85. IDENTIFY: Apply m = F a ! ! to each block. Parts (a) and (b) will be done together. Figure 5.85a Note that each block has the same magnitude of acceleration, but in different directions. For each block let the direction of a ! be a positive coordinate direction. SET UP: The free-body diagram for block A is given in Figure 5.85b. EXECUTE: y y a = AB A A Tm g m a −= () AB A a g = + 4.00 kg(2.00 m/s 9.80 m/s ) 47.2 N AB T = Figure 5.85b SET UP: The free-body diagram for block B is given in Figure 5.85b. EXECUTE: y y a = 0 B nmg = B nm g = Figure 5.85c 2 kk k (0.25)(12.0 kg)(9.80 m/s ) 29.4 N B fn m g μμ = = x x a = k B CA B B TT f m a = 2 k 47.2 N 29.4 N (12.0 kg)(2.00 m/s ) BC AB B TT f m a + = + + 100.6 N BC T =

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Applying Newton&s Laws 5-39 SET UP: The free-body diagram for block C is sketched in Figure 5.85d. EXECUTE: y y Fm a = CB C C mg T ma −= () C mga T 22 100.6 N 12.9 kg 9.80 m/s 2.00 m/s BC C T m ga == = −− Figure 5.85d EVALUATE: If all three blocks are considered together as a single object and m = F a ! ! is applied to this combined object, k . CA B A B C mg mg mg m m m a μ =+ + Using the values for k , A m and B m given in the problem and the mass C m we calculated, this equation gives 2 2.00 m/s , a = which checks. 5.86. IDENTIFY: Apply m = F a ! ! to each block. They have the same magnitude of acceleration, a . SET UP: Consider positive accelerations to be to the right (up and to the right for the left-hand block, down and to the right for the right-hand block). EXECUTE: (a) The forces along the inclines and the accelerations are related by (100 kg) sin30 (100 kg) and (50 kg) sin53 (50 kg) , Tg a g T a −° = ° = where T is the tension in the cord and a the mutual magnitude of acceleration. Adding these relations, (50 kg sin 53 100 kg sin 30 ) (50 kg 100 kg) , or 0.067 . g aa g °− ° = + =− Since a comes out negative, the blocks will slide to the left; the 100-kg block will slide down. Of course, if coordinates had been chosen so that positive accelerations were to the left, a would be 0.067 . g + (b) 2 2 0.067(9.80 m s ) 0.658 m s . a (c) Substituting the value of a (including the proper sign, depending on choice of coordinates) into either of the above relations involving T yields 424 N.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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166_PartUniversity Physics Solution - 5-38 Chapter 5 2 2 v...

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