Solutions to Homework Set #3 Phys2414 Fall 2005

Transcription

1 Solution Set #3 1 Solutions to Homework Set #3 Phys2414 Fall 2005 Note: The numbers in the boxes correspond to those that are enerated by WebAssin. The numbers on your individual assinment will vary. Any calculated quantities that involve these variable numbers will be boxed as well. 1. GRR1 3.P.001. Two cars, a Porsche and a Honda, are travelin in the same direction, althouh the Porsche is 186 m behind the Honda. The speed of the Porsche is 24.2 m/s and the speed of the Honda is 19.0 m/s. How much time does it take for the Porsche to catch the Honda? [Hint: What must be true about the displacements of the two cars when they meet?] If we call the time when the cars are 186 m apart t i, and the time when the Porshe catches the Honda t f, the equation of motion for the Porshe is iven by x f,p x f,p = v p (t f t i ) where x f,p and x i,p are the positions of the Porsche at times t f and t i, respectively, and v p is the speed of the Porshe. Likewise the equation of motion for the Honda is x f,h x i,h = v h (t f t i ) The hint for this problem wants us to reconize that at the time t f that the displacement of the cars is zero. In terms of our variables, this would be x f,h x f,p = 0 We can use this property if we subtract the equation of motion of the Porsche from the equation of motion of the Honda. Doin this ives us Rearranin the terms we et (x f,h x i,h ) (x f,p x i,p ) = [v h (t f t i )] [v p (t f t i )] (x f,h x f,p ) (x i,h x i,p ) = (v h v p ) (t f t i ) The first set of parentheses is the property we would like to invoke. Doin so and solvin for the chane in time we have (x i,h x i,p ) = (v h v p ) (t f t i ) x i,h x i,p v h v p = (t f t i ) Lettin t = (t f t i ) be the chane in time we have t = x i,h x i,p v h v p

2 Solution Set #3 2 The quantity in the numerator is the initial displacement, and the quantity in the denominator is the difference in speeds of the two cars. The chane in time is then t = 186 m 19.0 m/s 24.2 m/s t = 186 m 5.2 m/s t = 35.8 s 2. GRR1 3.P.020. In the fiure below (part a), two blocks are connected by a lihtweiht, flexible cord which passes over a frictionless pulley. (a) If = 3.2 k and = 5.4 k, what are the accelerations of each block? The net force on the two-block system is The acceleration of the blocks is then F net = + F net = ( + )a ( + )a = + ( ) m2 a = + ( ) 5.4 k 3.2 k a = 9.8 m/s k k a = 2.51 m/s 2 (b) What is the tension in the cord? The net force on the first block must be equal to the tension of the cord minus its weiht, i.e. F net(1) = T

3 Solution Set #3 3 The net force on the first block is also just its mass times the acceleration we found in part (a). Puttin this information toether we et Solvin for the tension ives us a = T T = (a + ) T = 3.2 k ( 2.5 m/s m/s 2) T = 39.4 N 3. GRR1 3.P.036. The minimum stoppin distance of a car movin at 38.6 mi/h is 12 m. Under the same conditions (so that the maximum brakin force is the same), what is the minimum stoppin distance for 56.4 mi/h? Work by proportions to avoid convertin units. The kinematic equation that relates distance, velocity, and acceleration is v 2 f v 2 i = 2a(x f x i ) where x i and v i are the initial position and velocity and x f and v f are the final postion and velocity with a bein the constant acceleration. Solvin for the acceleration we et 2a = v2 f v 2 i x f x i Since the maximum brakin force is the same for both stops, we know that the riht side of the above equation must be true for both stops. This means that v 2 a f v 2 a i x af x ai = v2 b f v 2 b i x bf x bi where the subscript a indicates the first stop, and b indicates the second stop. Lettin each of the stoppin distances equal a x and solvin for the stoppin distance of the second case, we et v 2 a f v 2 a i x a = v2 b f v 2 b i x b x b = v2 b f v 2 b i v 2 a f v 2 a i x a The stoppin distance in the second case is then x b = (0)2 ( 56.4 mi/h) 2 ( 12 m) (0) 2 ( 38.6 mi/h) 2 x b = 25.6 m

4 Solution Set # GRR1 3.P.067. In the fiure below (part a), two blocks are connected by a lihtweiht, flexible cord which passes over a frictionless pulley. If is 3.6 k and is 9.0 k, and block 2 is initially at rest 130 cm above the floor, how lon does it take block 2 to reach the floor? From proble, we know that the acceleration of the blocks is iven by ( ) m2 a = + The equation of motion of an object startin at rest is x = 1 2 at2 + x 0 Takin x x 0 = x, substitutin in the equation for the acceleration, and solvin for time we et x = 1 2 at2 x = 1 ( ) m2 t ( ) t 2 m2 + 1 = 2 x ( ) m2 + 1 t = 2 x ( ) t = 9.0 k k 1 2( 1.30 m) 9.0 k 3.6 k 9.8 m/s 2 t = s

5 Solution Set # GRR1 3.TB.016. The fiure below shows a plot of v x (t) for a car. (a) What is a av,x between t = 6 s and t = 11 s? (b) What is v av,x for the same time interval? (c) How far does the car travel from time t = 10 s to time t = 15 s? (a) The acceleration of the car, a av,x, is the chane in velocity per unit time. In terms of an equation this means or In this case the acceleration is then a av,x = v x t a av,x = v f,x v i,x t f t i a av,x = 14 m/s 4 m/s 11 s 6 s a av,x = 2 m/s 2 (b) Since the relationship between the velocities between times t = 6 s and t = 11 s is linear, the averae velocity between these two times is just the averae of the two velocities at those times. That is, The averae velocity is then v av,x = 1 2 (v f + v i ) v av,x = 1 (14 m/s + 4 m/s) 2 v av,x = 9 m/s (c) Since the velocity chanes between t = 10 s and t = 11 s we must be careful with our calculation. The chane in position is just the chane in time time the averae velocity durin that time, i.e. x = v av,x t

6 Solution Set #3 6 To find the distance travelled between the times t = 10 s and t = 11 s, we may just use the averae velocity durin that time. Therefore the total distance travelled is then x = 13 m/s (11 s 10 s) + 14 m/s (15 s 11 s) x = 69 m 6. GRR1 3.TB.028. A crate of oranes weihin 180 N rests on a flatbed truck 2.0 m from the back of the truck. The coefficients of friction between the crate and the bed are µ s = 0.30 and µ k = The truck drives on a straiht level hihway at a constant 8.0 m/s. (a) What is the force of friction actin on the crate? (b) If the truck speeds up with an acceleration of 1.0 m/s 2, what is the force of friction on the crate? (c) What is the maximum acceleration the truck can have without the crate startin to slide? (a) Since the truck is not acceleratin, there is no external force on the crate in the horizontal direction. Therefore there is no force of friction actin on the crate. F fr = 0 N. (b) If the truck accelerates at a truck = 1.0 m/s 2, the force on the crate in the horizontal direction is then F h = m crate a truck Assumin that the crate does not slip on the truck s flatbed, the force of static friction must be equal to this horizontal force, i.e. F fr = F h We can et the mass of the crate from the weiht iven in the problem. Therefore the force of friction on the crate is F fr = 180 N 1.0 m/s2 9.8 m/s2 F fr = 18 N (c) The maximum acceleration is found by settin F fr equal to the maximum force of friction. F fr,max = µ s F N where F N is the normal force on the crate. Settin this equal to the horizontal force and solvin for the acceleration ives The maximum acceleration is then F h = F fr,max m crate a truck = µ s F N a truck = µ s F N m crate a truck = µ s m crate m crate a truck = µ s a truck = 0.30(9.8 m/s 2 ) a truck = 2.9 m/s 2

7 Solution Set #3 7 Notice that the maximum acceleration does not depend on the mass of the crate, and it depends just on the coefficient of static friction. 7. GRR1 3.TB.032. A train is travelin south at 24.0 m/s when the brakes are applied. It slows down at a constant rate to a speed of 6.00 m/s in a time of 9.00 s. (a) What is the acceleration of the train durin the 9.00 s interval? (b) How far does the train travel durin the 9.00 s? (a) The acceleration is the chane in velocity per unit time, i.e. a = v t a = v f v i t The acceleration is then a = 6.00 m/s 24.0 m/s 9.00 s a = 2.00 m/s 2 Since the train is travelin south the neative sin we ot must mean that the acceleration is north. a = 2.00 m/s 2 to the north (b) The chane in position is then iven by x = v ave t where v ave is the ave velocity durin the acceleration. The chane in position is then x = x = ( 1 2 (v f + v i )) t ( ) 1 2 (24.0 m/s m/s) 9.00 s x = 15.0 m/s 9.00 s x = 135. m 8. GRR1 3.TB.044. A student, lookin toward his fourth-floor dormitory window, sees a flower pot with nasturtiums (oriinally on a window sill above) pass his 2.0-m-hih window in s. The distance between floors in the dormitory is 4.0 m. From a window on which floor did the flower pot fall? The flower pot, after bein released from a hiher floor, is accelerated by ravity. velocity of the flower pot as a function of time (assumin it starts at rest) is then The v y = t

8 Solution Set #3 8 At some time t w it passed by the student s window. The velocity at this time was v w,y = t w As measured by the student, the averae velocity of the flower pot as it passed his window was v w,y = h t where h is the heiht of the window and t is the time it took to pass it. Combinin this equation with the above expression and solvin for t w we et v w,y = t w y t = t w t w = h t The equation of motion for the flower pot as it falls under ravity is just y = 1 2 t2 The distance the flower pot fell to the window is then The flower pot then fell y w = 1 2 t2 w y w = 1 ( 2 h ) 2 t y w = 1 h 2 2 t 2 y w = 1 (2.0 m) m/s 2 (0.093 s) 2 y w = 24 m Given that each floor is 4.0 m tall, the flower pot must have fallen from 6 floors above the student. Since the student is on the fourth floor, the flower pot fell from the tenth floor.

9 Solution Set # GRR1 3.TB.074. The raph in the fiure below is the vertical velocity component v y of a bouncin ball as a function of time. The y-axis points up. Answer the followin based on the data in the raph. (a) At what time does the ball reach its maximum heiht? (b) For how lon is the ball in contact with the floor? (c) What is the maximum heiht of the ball? (a) Since a postive velocity indicates a ball movin upward, and a neative velocity indicates a ball movin downward, the ball must reach the apex when its velocity is zero. Therefore the time when it reaches its maximum heiht occurs first at t 0.3 s. (b) The ball is in contact with the floor from the time the velocity is neative to when it is positive aain. From the raph the ball must be in contact with the floor for about half of a tenth of a second 0.05 s. (Lookin riht after 2.5 s helped me determine this.) (c) The maximum heiht of the ball may be determined by takin the averae of the velocities between when the ball first moves upward and when it is at the apex and multiplyin by the time it takes to do that. The ball starts upward with a velocity of 3.0 m/s. We know from part (a) that it reaches the apex in about 0.3 s. The maximum heiht is then y = v ave t ( ) 1 y = 2 (3.0 m/s + 0 m/s) 0.3 s y = 0.45 m 10. GRR1 3.TB.075. [219493] A rocket enine can accelerate a rocket launched from rest vertically up with an acceleration of 20.0 m/s 2. However, after 50.0 seconds of fliht the enine fails. Nelect air resistance. (a) What is the rocket s elevation when the enine fails? (b) When does it reach its maximum heiht? (c) What is the maximum heiht reached? [Hint: a raphical solution may be easiest.] (a) The equation of motion for the rocket is iven by y = 1 2 a eninet 2 + y 0

10 Solution Set #3 10 where a enine is the accelereation due to the enine, and y 0 is the initial heiht. Assumin the rocket is launched from sea level, the elevation when the rocket fails is y f = 1 2 a enine t 2 enine y f = m/s2 (50.0 s) 2 y f = 10.0 m/s s 2 y f = m y f = 25.0 km (b) After the enine fails, the rocket is simply under the influence of ravity. Therefore it will undero constant acceleration, but downward this time. The velocity as a function of time for an object under constant acceleration is iven by v y = a y t + v f,y In this case the acceleration, a y, is that due to ravity, and the initial velocity, v 0,y, is the velocity the rocket had reached when the enine failed. The velocity that the rocket had reached is iven by v f,y = a enine t enine where t enine is the time the rocket enine was on. The equation for the velocity after the enine fails then becomes v y = t + a enine t enine From Problem 9, we know that the rocket will have reached its maximum heiht when the vertical velocity is equal to zero. Settin the equation for the velocity equal to zero and solvin for the time ives 0 = t apex + a enine t enine t apex = a enine t enine t apex = a enine t enine The time when the rocket reaches its maximum heiht after the enine fails is then t apex = 20.0 m/s2 9.8 m/s 50.0 s 2 t apex = 102 s Therefore the rocket reaches its maximum heiht 102 s s = 152 s into the fliht. (c) After the enine fails the equation of motion for the rocket is y = 1 2 t2 + v f,y t + y f

11 Solution Set #3 11 where v f,y and y f are the velocity and elevation it reached when the enine failed. Substitutin in these quantities, which we found in parts (a) and (b), we et y apex = 1 2 t2 apex + v f,y t apex + y f y apex = 1 ( ) 2 ( ) 2 aenine aenine t enine + (a enine t enine ) t enine a enine t 2 enine y apex = 1 (a enine t enine ) 2 2 (a enine t enine ) 2 y apex = 1 2 y apex = 1 2 a enine t 2 enine + (a enine t enine ) a enine t 2 enine ) ( aenine a enine t 2 enine The maximum heiht the rocket reaches is then y apex = 1 ( ) 20.0 m/s m/s2 (50.0 s) m/s y apex = m y apex = 76.0 km The rocket reaches a maximum heiht of 76.0 km. 11. GRR1 3.TB.082. Two blocks, masses and, are connected by a massless cord (the fiure below ). If the two blocks are accelerated uniformly on a frictionless surface by applyin a force of manitude T 2 to a second cord connected to, what is the ratio of the tensions in the two cords in terms of the masses? T 1 /T 2 =? The net force on the two-block system is F net = ( + )a where a is the acceleration. The tension in the second rope must be equal to this net force. T 2 = F net T 2 = ( + )a The tension in the first rope is equal to the force required to accelerate the first block at an acceleration of a, i.e. T 1 = a

12 Solution Set #3 12 Therefore the ratio of the tensions is iven by dividin the two equations for each of the tensions T 1 T 2 = T 1 T 2 = a ( + )a GRR1 3.TB.062. In the fiure below (part a), the block of mass slides to the riht with coefficient of kinetic friction µ k on a horizontal surface. The block is connected to a hanin block of mass by a liht cord that passes over a liht, frictionless pulley. (a) Find the acceleration of each of the blocks and the tension in the cord. (b) Check your answers in the special cases,, and m1 = m2. (c) For what value of (if any) do the two blocks slide at constant velocity? What is the tension in the cord in that case? (a) The net force on the two-block system is iven by F net = F fr + F W where F fr is the force of friction on the first block, and F W is the weiht of the second block. Since F fr opposes F W, we will call rihtward and downward positive. The acceleration of the blocks is then ( + )a = µ k + ( + )a = ( µ k ) a = µ k + The net force on the first block is equal to the tension plus the force of friction F net,1 = T + F fr F net,1 = T F fr a = T µ k

13 Solution Set #3 13 Solvin for the tension we et T = a + µ k T = (a + µ k ) Now we can substitute in what we found the acceleration to be ( ) m2 µ k T = + µ k + Simplifyin this we et ( m2 µ k T = + µ ) k( + ) + + ( m2 µ k + µ k + µ k T = T = ( m2 + µ k + T = (1 + µ k ) + + m ) 1 (b) To ensure that our answers make sense, we consider the cases when the first block is really liht, when it is really heavy, and when the masses are equal. To say that the mass of the first block is really liht compared to the second block mathematically, we write or 1 This means when we encounter the ratio / we may treat it as zero. We can find this ratio in our expression for the acceleration if we rearrane a bit, a = µ k + a = µ k + + a = (1 + ) µ k (1 + a = (1 + ) µ m1 k (1 + ) Now if we treat every instance of / as zero we et a (1 + 0) µ k 0 (1 + 0) a 0 a ) )

14 Solution Set #3 14 This result makes ood sense because if the first block were very liht, the second block would fall as if it were not there. Similarly, the tension can be rewritten as T = (1 + µ k ) + T = (1 + µ k ) (1 + ) T = (1 + µ k ) (1 + ) T (1 + µ k )(0) T 0 The tension would be very small if the mass of the first block was very small. For the case of the first block bein much heavier than the second block, we look for the ratio / and treat it as zero. The acceleration then becomes a = µ k + a = µ k + + a = ( + 1) µ k ( + 1) a = m µ k ( + 1) a (0) (0) + 1 µ k ((0) + 1) a µ k a µ k This result is somewhat nonsense. The first block would not accelerate to the left if it was very heavy, but it does point out that the force of friction dominates over the weiht of the second block. It that sense, it does have meanin. So if the first block has a lare mass compared to the second block, the acceleration is effectively zero. The tension would then act like T = (1 + µ k ) + T = (1 + µ k ) ( + 1) T = (1 + µ k ) + 1 T (1 + µ k ) (0) + 1 T (1 + µ k )

15 Solution Set #3 15 It makes ood sense that the tension in the cord just becomes the weiht of the second block, because the two-block system should not move if the first block is lare enouh. For the case of the masses bein equal, if we let = = m, the acceleration becomes a = m µ km m + m a = (1 µ k ) m a = 1 2 (1 µ k) We can see that if µ k were small the acceleration would be like 1 or half of the acceleration 2 due to ravity. The tension would become T = (1 + µ k ) m m m + m T = (1 + µ k ) m2 2m T = 1 2 m(1 + µ k) Aain, if the coefficient of friction were small, the tension would just be half the weiht of one of the blocks. (c) The blocks slide at constant velocity when the net force is zero or when the acceleration is zero. Settin the acceleration equal to zero and solvin for we et 0 = µ k + 0 = µ k + + = µ k + + = µ k = µ k Therefore the blocks would slide at a constant velocity if = µ k. The tension in the cord for this value of is then T = (1 + µ k ) (µ k ) (µ k ) + T = (1 + µ k ) 1µ k (µ k + 1) T = µ k