acceleration of 2.4 m/s. (b) Now, we have two rubber bands (force 2F) pulling two glued objects (mass 2m). Using F ma, 2.0 furlongs x 2.0 s 2 4.

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1 Model: An object s acceleration is linearl proportional to the net force. Solve: (a) One rubber band produces a force F, two rubber bands produce a force F, and so on. Because F a and two rubber bands (force F) produce an acceleration of. /s, four rubber bands will produce an acceleration of.4 /s. (b) Now, we have two rubber bands (force F) pulling two glued objects (ass ). Using F a, F ( ) a a F/ 0.6 /s 5.0 Solve: Use proportional reasoning. Given that distance traveled is proportional to the square of the d tie, d t, so should be constant. We have t.0 furlongs.0 s 4.0 s Thus the distance traveled b the object in 4.0 s is = 8.0 furlongs. Assess: A longer tie should result in a longer distance traveled. 5.4 Solve: Newton s second law tells us that F a. Copute F for each case: (a) F (0.00 kg)(5 /s ) N. (b) F (0.00 kg)(0 /s ) N. Assess: To double the acceleration we ust double the force, as epected Solve: The object will be in equilibriu if F 3 has the sae agnitude as F F but is in the opposite direction so that the su of all the three forces is zero. 5..

2 Solve: The free-bod diagra shows three forces with a net force (and therefore net acceleration) upward. There is a force labeled F G directed down, a force Fthrust directed up, and a force D directed down. So a possible description is: A rocket accelerates upward Assess: Since the velocit is constant, the acceleration is zero, and the net force is zero Solve: According to Newton s second law F a, the force at an tie is found sipl b ultipling the value of the acceleration b the ass of the object Model: Use the particle odel for the object. Solve: (a) We are told that for an unknown force (call it F 0) acting on an unknown ass (call it 0) the acceleration of the ass is becoes F 0. Newton s second law gives 0 /s. According to Newton s second law, F a s This eans a is 5 /s. (b) The force is F 0 and the ass is now 0. Newton s second law gives This eans a 0 /s. (c) A siilar procedure gives (d) A siilar procedure gives a 0 /s. a.5 /s. F a s F 0 0 (0 /s ). The force then

3 5.38. Solve: (d) This is an object on a surface because FG n. It ust be oving to the left because the kinetic friction is to the right. The description of the free-bod diagra could be: A copressed spring is shooting a plastic block to the left Solve: (d) There is a thrust at an angle to the horizontal and a gravitational force. There is no noral force so the object is not on a surface. The description could be: A rocket is fired at an angle to the horizontal and there is no drag force The noral force is perpendicular to the hill. The frictional force is parallel to the hill The ball rests on the floor of the barrel because the gravitational force is equal to the noral force. There is a force of the spring to the right which causes an acceleration.

4 6.. Model: We can assue that the ring is a particle. This is a static equilibriu proble. We will ignore the weight of the ring, because it is ver light, so the onl three forces are the tension forces shown in the free-bod diagra. Note that the diagra defines the angle. Solve: Because the ring is in equilibriu it ust obe Fnet 0 N. This is a vector equation, so it has both - and -coponents: F T cos T 0 N T3cos T net 3 F T T sin 0 N T sin T net 3 3 We have two equations in the two unknowns T 3 and. Divide the -equation b the -equation: Now we can use the -equation to find T3sin T 80 N tan.6 tan.6 58 T cos T 50 N 3 T T cos 50 N 94 N cos58 3 The tension in the third rope is 94 N directed 58 below the horizontal Please refer to Figure EX6.6. Solve: For the diagra on the left, three of the vectors lie along the aes of the tilted coordinate sste. Notice that the angle between the 3 N force and the -ais is the sae 0 b which the coordinates are tilted. Appling Newton s second law, F 5 N N 3sin 0 N a.49 /s kg net F net.8 N 3cos0 N 0 /s a kg For the diagra on the right, the -newton force in the first quadrant akes an angle of 5 with the positive - ais. The other -newton force akes an angle of 5 with the negative -ais. The accelerations are F cos5 N sin5 N 3 N a 0.8 /s kg net F.44 N sin5 N cos5 N a 0 /s kg net

5 6.0. Model: We assue that the bo is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the onl horizontal force. Solve: (a) Since the bo is at rest, a 0 /s, and the net force on the bo ust be zero. Therefore, according to Newton s first law, the tension in the rope ust be zero. (b) For this situation again, a 0 /s, so Fnet T 0 N. (c) Here, the velocit of the bo is irrelevant, since onl a change in velocit requires a nonzero net force. Since a 5.0 /s, Fnet T a 50 kg 5.0 /s 50 N Assess: For parts (a) and (b), the zero acceleration iediatel iplies that the rope is eerting no horizontal force on the bo. For part (c), the 50 N force (the equivalent of about half the weight of a sall person) sees reasonable to accelerate a bo of this ass at 5.0 /s Model: Use the particle odel for the woan. Solve: (a) The woan s weight on the earth is wearth gearth 55 kg 9.80 /s 540 N (b) Since ass is a easure of the aount of atter, the woan s ass is the sae on the oon as on the earth. Her weight on the oon is woon goon 55 kg.6 /s 89 N Assess: The saller acceleration due to gravit on the oon reveals that objects are less strongl attracted to the oon than to the earth. Thus the woan s saller weight on the oon akes sense Model: We assue that the ule is a particle acted on b two opposing forces in a single line: the farer s pull and friction. The ule will be subject to static friction until (and if!) it begins to ove; after that it will be subject to kinetic friction.

6 Solve: Since the ule does not accelerate in the vertical direction, the free-bod diagra shows that n F g The aiu friction force is G. fsa sg kg 9.80 /s 940 N The aiu static friction force is greater than the farer s aiu pull of 800 N; thus, the farer will not be able to budge the ule. Assess: The farer should have known better. 6.. Model: We assue that the plane is a particle accelerating in a straight line under the influence of two forces: the thrust of its engines and the rolling friction of the wheels on the runwa. We can use one-diensional kineatics. Solve: We can use the definition of acceleration to find a, and then appl Newton s second law. We obtain: v 8 /s 0 /s a.34 /s t 35 s F F F f a F f a net thrust r thrust r For rubber rolling on concrete, r 0.0 (Table 6.), and since the runwa is horizontal, n FG g. Thus: F F a g a g a thrust r G r r 75,000 kg /s.34 /s 90,000 N Assess: It s hard to evaluate such an enorous thrust, but coparison with the plane s ass suggests that 90,000 N is enough to produce the required acceleration.

7 6.6. Model: We will represent the tennis ball as a particle. The tennis ball falls straight down toward the earth s surface. The ball is subject to a net force that is the resultant of the gravitational and drag force vectors acting verticall, in the downward and upward directions, respectivel. Once the net force acting on the ball becoes zero, the terinal velocit is reached and reains constant for the rest of the otion. Solve: The atheatical equation defining the dnaical equilibriu situation for the falling ball is Fnet FG D 0 N Since onl the vertical direction atters, one can write: F 0 N F D F 0 N net When this condition is satisfied, the speed of the ball becoes the constant terinal speed v v. The ter agnitudes of the gravitational and drag forces acting on the ball are: F g 9.80 /s G D Avter 0.5R vter /s 0.56 N 4 The condition for dnaic equilibriu becoes: 0.56 N 9.80 /s 0.56 N 0 N 57 g 9.80 /s Assess: The value of the ass of the tennis ball obtained above sees reasonable Model: The plastic ball is represented as a particle in static equilibriu. G Solve: The electric force, like the weight, is a long-range force. So the ball eperiences the contact force of the string s tension plus two long-range forces. The equilibriu condition is F T F T sin F 0 N net elec elec F T F T cos g 0 N net G

8 We can solve the -equation to get Substituting this value into the -equation, 0.00 kg9.8 /s g T N cos cos0 F T (b) The tension in the string is N. elec sin.04 0 N sin N Model: We can assue our bod is a particle oving in a straight line under the influence of two forces: gravit and the support force of the scale. Solve: The weight (see Equation 6.0) of an object oving in an elevator is When accelerating upward, the acceleration is When braking, the acceleration is a w w g a g g g a 70 lb 9.80 /s.3 /s 50 lb a 0 lb 9.80 /s.0 /s 50 lb Assess: A 0-0% change in apparent weight sees reasonable for a fast elevator, as the ones in the Epire State Building ust be. Also note that we did not have to convert the units of the weights fro pounds to newtons because the weights appear as a ratio.

9 6.38. Model: We will represent the bullet as a particle. Solve: (a) We have enough inforation to use kineatics to find the acceleration of the bullet as it stops. Then we can relate the acceleration to the force with Newton s second law. (Note that the barrel length is not relevant to the proble.) The kineatic equation is /s v v a a v 400 /s 0. Notice that a is negative, in agreeent with the vector a in the otion diagra. Turning to forces, the wood eerts two forces on the bullet. First, an upward noral force that keeps the bullet fro falling through the wood. Second, a retarding frictional force f k that stops the bullet. The onl horizontal force is f, k which points to the left and thus has a negative -coponent. The -coponent of Newton s second law is 5 F f a f a net k k 0.0 kg /s 6670 N Notice how the signs worked together to give a positive value of the agnitude of the force. (b) The tie to stop is found fro v v0 a t as follows: (c) v s 600 μs t a Using the above kineatic equation, we can find the velocit as a function of t. For eaple at v /s ( /s )(60 0 s) 360 /s t 60 μs,

10 6.4. Model: We assue Sa is a particle oving in a straight line down the slope under the influence of gravit, the thrust of his jet skis, and the resisting force of friction on the snow. Solve: Fro the height of the slope and its angle, we can calculate its length: h h 50 sin 0 88 sin sin0 0 Since Sa is not accelerating in the -direction, we can use Newton s first law to calculate the noral force: F cos 0 N net F n FG n F G cos g cos (75 kg)(9.80 /s )(cos0 ) 74 N One-diensional kineatics gives us Sa s acceleration: v v a 0 0 Then, fro Newton s second law and the equation fk k n: v 40 /s 0 /s v 0 a.78 /s 88 F F F sin F f a net G thrust k g sin Fthrust a k n 75 kg9.80 /s sin0 00 N 75 kg.78 /s N Assess: This coefficient sees a bit high for skis on snow, but not ipossible.

11 6.46. Model: We will represent the wood block as a particle, and use the odel of kinetic friction and kineatics. Assue w sin f, s so it does not hang up at the top. The block ends where it starts, so 0 0. We epect v to be negative, because the block will be oving in the -direction, so we ll want to take v as the final speed. Because of friction, we epect to find v v. 0 Solve: (a) The friction force is opposite to v, and up the slope during the second half. upward otion is so f k points down the slope during the first half of the otion F G and n are the onl other forces. Newton s second law for the Fnet FG sin fk g sin fk a a0 Fnet n FG cos n g cos a 0 /s The friction odel is f k k n. First solve the -equation to give n g cos. Use this in the friction odel to get fk kg cos. Now substitute this result for f k into the -equation: g sin kg cos a0 gsin kcos 9.8 /s sin30 0.0cos /s Kineatics now gives v v 0 /s 0 /s a /s v v a 7.6 The block s height is then h sin (7.6 )sin (b) For the return trip, f k points up the slope, so the -coponent of the second law is F F sin f g sin f a a net G k k Note the sign change. The -equation and the friction odel are unchanged, so we have The kineatics for the return trip are sin cos 3.0 /s a g k

12 v v a v a 3.0 /s /s Notice that we used the negative square root because v is a velocit with the vector pointing in the -direction. The final speed is v 7.0 /s Model: The antiques (ass ) in the back of our pickup (ass M) will be treated as a particle. The antiques touch the truck s steel bed, so onl the steel bed can eert contact forces on the antiques. The pickup-antiques sste will also be treated as a particle, and the contact force on this particle will be due to the road. Solve: (a) We will find the sallest coefficient of friction that allows the truck to stop in 55, then copare that to the known coefficients for rubber on concrete. For the pickup-antiques sste, with ass M, Newton s second law is ( F ) F N F ( f ) 0 N 0 N f ( M) a ( M) a net G PA net G PA ( F ) F N F ( f ) N ( M ) g 0 N 0 N The odel of static friction is f N, where is the coefficient of friction between the tires and the road. These equations can be cobined to ield ag. Since constant-acceleration kineatics gives v v a( ), we find 0 0 g 5 /s v v v a /s in 0 0

13 The truck cannot stop if is saller than this. But both the static and kinetic coefficients of friction,.00 and 0.80 respectivel (see Table 6.), are larger. So the truck can stop. (b) The analsis of the pickup-antiques sste applies to the antiques, and it gives the sae value of 0.58 for. This value is saller than the given coefficient of static friction in ( s 0.60) between the antiques and the truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55. Assess: The analsis of parts (a) and (b) are the sae because ass cancels out of the calculations. According to the California Highwa Patrol Web site, the stopping distance (with zero reaction tie) for a passenger vehicle traveling at 5 /s or 8 ft/s is approiatel 43. This is saller than the 55 over which ou are asked to stop the truck Model: The ball is a particle eperiencing a drag force and traveling at twice its terinal velocit. Solve: (a) An object falling at greater than its terinal velocit will slow down to its terinal velocit. Thus the drag force is greater than the force of gravit, as shown in the free-bod diagras. When the ball is shot straight up, 4g F a F D g Av g A v g A 5g 4 4 A G ter Thus a 5 g, where the inus sign indicates the downward direction. We have used Equations 6.6 for the drag force and 6.9 for the terinal velocit. (b) When the ball is shot straight down, Thus a 3 g, this tie directed upward. (c) F a D F A v g 3g 4 G ter The ball will slow down to its terinal velocit, slowing quickl at first, and ore slowl as it gets closer to the terinal velocit because the drag force decreases as the ball slows.

14 6.58. Model: The ball hanging fro the ceiling of the truck b a string is represented as a particle. Solve: (a) You cannot tell fro within the truck. Newton s first law sas that there is no distinction between at rest and constant velocit. In both cases, the net force acting on the ball is zero and the ball hangs straight down. (b) Now ou can tell. If the truck is accelerating, then the ball is tilted back at an angle. (c) The ball oves with the truck, so its acceleration is 5 /s in the forward direction. (d) The free-bod diagra shows that the horizontal coponent of T provides a net force in the forward direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newton s second law for the ball is sin0 Fnet T FG Fnet T T T cos0 g a a 0 /s We can solve the second equation for the agnitude of the tension:.0 kg9.8 /s g T 9.95 N cos0 cos0 Then the first equation gives the acceleration of the ball and truck: The truck s velocit cannot be deterined. T sin0 a 9.95 Nsin0 =.73 /s 6.6. Solve: (a) The terinal velocit for a falling object is reached when the downward gravitational force is balanced b the upward drag force. (b) The ass of the spherical sand grain of densit FG ter D g bv 6Rv v gr ter g 6R ter 3 p 400 kg/ is R kg/ 9.80 /s Thus vter.3 /s Ns The tie required for the sand grain to fall 50 at this speed is t 38 s..3 /s Assess: The speed of.3 /s for a sand grain falling through water sees about right.

15 6.66. Solve: (a) A driver traveling at 40 /s in her 500 kg auto slas on the brakes and skids to rest. How far does the auto slide before coing to rest? (b) (c) Newton s second law is F n FG n g a 0 N F 0.80n a The -coponent equation gives n g ields 500 kg 9.8 /s. Substituting this into the -coponent equation 500 kga kg9.8 /s a Using the constant-acceleration kineatic equation v v0 a, we find 7.. v 40 /s 0 0 a 7.8 /s /s 7.8 /s Solve: (a) Both the bowling ball and the soccer ball have a noral force fro the surface and gravitational force on the. The interaction forces between the two are equal and opposite. (b) The sste consists of the soccer ball and bowling ball, as indicated in the figure.

16 (c) Assess: Even though the soccer ball bounces back ore than the bowling ball, the forces that each eerts on the other are part of an action/reaction pair, and therefore have equal agnitudes. Each ball s acceleration due to the forces on it is deterined b Newton s second law, a F /, net which depends on the ass. Since the asses of the balls are different, their accelerations are different Please refer to Figure EX7.6. Solve: (a) For each block, there is a gravitational force with the earth, a noral force and kinetic friction with the surface, and a tension force due to the rope. (b) The tension in the assless ropes over the frictionless pulle is the sae on both blocks. Block A accelerates down the incline with the sae acceleration that Block B has up the incline. The sste consists of the two blocks, as indicated in the figure. (c)

17 Assess: The inclined coordinate sstes allow the acceleration a to be purel along the -ais. This is convenient since then one coponent of a is zero, siplifing the atheatical epression of Newton s second law Model: The blocks are to be odeled as particles and denoted as,, and 3. The surface is frictionless and along with the earth it is a part of the environent. The three blocks are our three sstes of interest. The force applied on block is FA on N. The acceleration for all the blocks is the sae and is denoted b a. Solve: (a) Newton s second law for the three blocks along the -direction is Fon FA on Fon a Fon Fon F3 on a F F a on 3 on 3 3 Adding these three equations and using Newton s third law ( F on F on and F3 on F on 3), we get F a A on 3 Using this value of a, the force equation on block 3 gives (b) Substituting into the force equation on block, N kg kg 3 kg a a /s F on 3 3a 3 kg /s 6 N on N F kg /s F on 0 N Assess: Because all three blocks are pushed forward b a force of N, the value of 0 N for the force that the kg block eerts on the kg block is reasonable.

18 7.4. Model: The block of ice (I) is a particle and so is the rope (R) because it is not assless. We ust therefore consider both the block of ice and the rope as objects in the sste. Solve: The force F et acts onl on the rope. Since the rope and the ice block ove together, the have the sae acceleration. Also because the rope has ass, F et on the front end of the rope is not the sae as FI on R that acts on the rear end of the rope. Newton s second law along the -ais for the ice block and the rope is 7.8. Please refer to Figure P7.8. F F a on I R on I I 0 kg.0 /s 0 N Fon R F et FI on R Ra et R on I R F F a Fet FR on I Ra 0 N kg.0 /s N Solve: Since the ropes are assless we can treat the tension force the transit as a Newton s third law force pair on the blocks. The connection shown in figure P7.8 has the sae effect as a frictionless pulle on these assless ropes. The blocks are in equilibriu as the ass of A is increased until block B slides, which occurs when the static friction on B is at its aiu value. Appling Newton s first law to the vertical forces on block n F g The static friction force on B is thus B gives B. G B B f n g s B s B s B. Appling Newton s first law to the horizontal forces on B gives f vertical forces on A gives T F g Since T, A on B B on A B on A G A A. g g s B A s T B A on B, T we have f A sb kg.0 kg and the sae analsis of the s B Ag, so

19 7.. Consider a segent of the rope of length, starting fro the botto of the rope. The weight of this segent of rope is a downward force. It is balanced b the tension force at height. Solve: The ass of this segent of rope is the sae fraction of the total ass M. kg as length is a fraction of the total length L 3.0. That is, / M / L, fro which we can write the ass of the rope segent M L This segent of rope is in static equilibriu, so the tension force pulling up on it is Mg (. kg)(9.8 /s ) T FG g 7.9 N L 3.0 where is in. The tension increases linearl fro 0 N at the botto ( 0 ) to.6 N at the top ( 3 ). This is shown in the graph Model: The starship and the shuttlecraft will be denoted as M and, respectivel, and both will be treated as particles. We will also use the constant-acceleration kineatic equations. Solve: (a) The tractor bea is soe kind of long-range force F. M on Regardless of what kind of force it is, b Newton s third law there ust be a reaction force F on M on the starship. As a result, both the shuttlecraft and the starship ove toward each other (rather than the starship reaining at rest as it pulls the shuttlecraft in). However, the ver different asses of the two crafts eans that the distances the each ove will also be ver different. The pictorial representation shows that the eet at tie t when. M There s onl one force on each craft, so Newton s second law is ver siple. Furtherore, because the forces are an action/reaction pair,

20 The accelerations of the two craft are a M F F F F 4.00 N 0.00 /s M.00 kg 4 M on on M tractor bea N 4 on M 6 a F N.00 kg 4 M on.0 /s 4 Acceleration a is negative because the force and acceleration vectors point in the negative -direction. The accelerations are ver different even though the forces are the sae. Now we have a constant-acceleration proble in kineatics. At a later tie t the positions of the crafts are The craft eet when, so M a t a t M 0 v t t a t t a t M M0 M0 0 M 0 M v t t a t t a t ,000 t 99.5 s 0 0 am a am a.0 /s Knowing t, we can now find the starship s position as it eets the shuttlecraft: a t 99 M M The starship oves 99 as it pulls in the shuttlecraft fro 0 k awa Model: The two blocks for a sste of interacting objects. Please refer to Figure P7.30. Solve: It is possible that the left-hand block (Block L) is accelerating down the slope faster than the right-hand block (Block R), causing the string to be slack (zero tension). If that were the case, we would get a zero or negative answer for the tension in the string. Newton s first law applied to the -direction on Block L ields Therefore F 0 n F cos0 n g cos0 L L G L L L fk k Lg cos kg 9.80 /s cos0.84 N L L A siilar analsis of the vertical forces on Block R gives fk.84 N as well. Using Newton s second law in R the -direction for Block L, F L La T R on L fk F G sin0 L L La TR on L.84 N Lg sin0. For Block R, FR Ra F sin0.84 N G T R L on R Ra R gsin0.84 N TL on R These are two equations in the two unknowns a and TL on R TR on L T. Solving the, we obtain T 0.6 N. a. /s and

21 Assess: The tension in the string is positive, and is about /3 of the kinetic friction force on each of the blocks, which is reasonable Model: Blocks and ake up the sste of interest and will be treated as particles. Assue a assless rope and frictionless pulle. Solve: The blocks accelerate with the sae agnitude but in opposite directions. Thus the acceleration constraint is a, a a where a will have a positive value. There are two real action/reaction pairs. The two tension forces will act as if the are action/reaction pairs because we are assuing a assless rope and a frictionless pulle. Make sure ou understand wh the friction forces point in the directions shown in the freebod diagras, especiall force f eerted on block b block. We have quite a few pieces of inforation to include. First, Newton s second law for blocks and : F f T n T a a net on k F n g 0 N n g net on F T f f T T f n T a a net on pull pull k F n n g 0 N n n g net on We ve alread used the kinetic friction odel in both -equations. Net, Newton s third law: n n g f f kn k g TT T Knowing n, we can now use the -equation of block to find n. Substitute all these pieces into the two - equations, and we end up with two equations in two unknowns: k g T a Subtract the first equation fro the second to get T T g g a pull k k T 3 g Tpull k 3 g a a.77 /s pull k

22 7.38. Model: Assue the particle odel for,, and, 3 and the odel of kinetic friction. Assue the ropes to be assless, and the pulles to be frictionless and assless. Solve: Newton s second law for is T ( FG ) a. Newton s second law for is Fon n FG n Newton s second law for 3 is 0 N.0 kg 9.8 /s 9.6 N ( F ) T f T a T n T (.0 kg) a on k k 3 F T F a on G Since,, and 3 ove together, a a a3 a. The equations for the three asses thus becoe T n T a a T FG a.0 kg a k.0 kg Subtracting the third equation fro the su of the first two equations ields: FG kn FG 6.0 kg a 3 T FG 3 3a 3.0 kg a.0 kg 9.8 /s N 3.0 kg 9.8 /s 6.0 kg a a.3 /s

23 7.4. Model: Use the particle odel for the cable car and the counterweight. Assue a assless cable. Solve: (a) Notice the separate coordinate sstes for the cable car (object ) and the counterweight (object ). Forces T and T act as if the are an action/reaction pair. The braking force F B works with the cable tension T to allow the cable car to descend at a constant speed. Constant speed eans dnaic equilibriu, so Fnet 0 N for both sstes. Newton s second law for the cable car is F T F gsin 0 N Fnet on n g cos 0 N net on B Newton s second law for the counterweight is F gsin T 0 N Fnet on n g cos 0 N net on Fro the -equation for the counterweight, T g sin. B Newton s third law, T. T Thus the -equation for the cable car becoes F g sin T g sin g sin 3770 N B (b) If the brakes fail, then FB 0 N. The car will accelerate down the hill on one side while the counterweight accelerates up the hill on the other side. Both will have negative accelerations because of the direction of the acceleration vectors. The constraint is a a a, where a will have a negative value. Using TT T, the two -equations are F T gsin a a net on F gsin T a a net on Note that the -equations aren t needed in this proble. Add the two equations to eliinate T: sin sin gsin g sin a a g 0.99 /s Now we have a proble in kineatics. The speed at the botto is calculated as follows: v a v v a a /s /s Assess: A speed of approiatel 60 ph as the cable car travels a distance of 000 along a frictionless slope of 30 is reasonable.

24 7.46. Model: Use the particle odel for the wedge and the block. The block will not slip relative to the wedge if the both have the sae horizontal acceleration a. Note: n n on on. Solve: The -coponent of Newton s second law for block is g cos Fon n on cos FG 0 N n on Cobining this equation with the -coponent of Newton s second law ields: Now, Newton s second law for the wedge is n sin F n a a g tan on on on sin F F n sin a on on F a n sin a a ( ) a ( ) g tan on