Answer, Key Homework 4 David McIntyre Mar 25,

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1 Answer, Key Homework 4 Dvid McIntyre Mr 25, his print-out should hve 18 questions. Multiple-choice questions my continue on the next column or pe find ll choices before mkin your selection. he due time is Centrl time. Chpters 3 nd 4 problems. 001 (prt 1 of 1) 0 points A bll on the end of strin is whirled round in horizontl circle of rdius m. he plne of the circle is 1.49 m bove the round. he strin breks nd the bll lnds 2.71 m wy from the point on the round directly beneth the bll s loction when the strin breks. he ccelertion of rvity is 9.8 m/s 2. Find the centripetl ccelertion of the bll durin its circulr motion. Correct nswer: m/s 2. In order to find the centripetl ccelertion of the bll, we need to find the initil velocity of the bll. Let y be the distnce bove the round. After the strin breks, the bll hs no initil velocity in the verticl direction, so the time spent in the ir my be deduced from the kinemtic eqution, Solvin for t, y 1 2 t2 t 2y Let d be the distnce trveled by the bll. hen v x d t d 2y Hence, the centripetl ccelertion of the bll durin its circulr motion is c v2 x r d2 2yr m/s2 002 (prt 1 of 2) 0 points he orbit of Moon bout its plnet is pproximtely circulr, with men rdius of m. It tkes 26.4 dys for the Moon to complete one revolution bout the plnet. Find the men orbitl speed of the Moon. Correct nswer: m/s. Dividin the lenth C 2πr of the trjectory of the Moon by the time 26.4 dys s of one revolution (in seconds!), we obtin tht the men orbitl speed of the Moon is v C 2 π r 2 π ( m ) s m/s. 003 (prt 2 of 2) 0 points Find the Moon s centripetl ccelertion. Correct nswer: m/s 2. Since the mnitude of the velocity is constnt, the tnentil ccelertion of the Moon is zero. For the centripetl ccelertion we use the formul c v2 r ( m/s ) m m/s (prt 1 of 3) 0 points A trin slows down t constnt rte s it rounds shrp circulr horizontl turn. Its initil speed is not known. It tkes 16.4 s to slow down from 78 km/h to 30 km/h. he rdius of the curve is 180 m. As the trin oes round the turn, wht is the mnitude of the tnentil component of the ccelertion? Correct nswer: m/s 2. Bsic concepts he components of norml nd tnentil ccelertion re iven by t v t

2 Answer, Key Homework 4 Dvid McIntyre Mr 25, [ ( )] r v r. (50 km/h) 3600 Solution: Since the tnentil ccelertion 180 m chnes the mnitude of the velocity nd the m/s 2. norml ccelertion chnes the direction of the velocity, with the iven initil nd finl velocities v i 78 km/h nd v f 30 km/h with herefore, the mnitude of the totl ccelertion is the time intervl t 16.4 s, the tnentil ccelertion is t v 2 t + 2 r t v f v i ( m/s 2 ) 2 + ( m/s 2 ) 2 [ t ] m/s 2. (78 km/h) (30 km/h) 16.4 s ( 1000 m 1 km m/s 2 t m/s 2. ) ( 1 h 3600 s 005 (prt 2 of 3) 0 points As the trin oes round the turn, wht is the sin of the tnentil component of the ccelertion? ke the direction in which the trin is movin to be the positive direction. 1. correct Not enouh informtion See previous solution. 006 (prt 3 of 3) 0 points At the moment the trin s speed is 50 km/h, wht is the mnitude of the totl ccelertion? Correct nswer: m/s 2. When v 50 km/h, r 180 m, the norml ccelertion is t v2 r ) 007 (prt 1 of 2) 5 points wo msses m 1 nd re connected in the mnner shown. m he system is ccelertin downwrd with ccelertion of mnitude 2. Determine 2. ( ) m ( m 1 ) ( ) 2 m correct 2 2

3 Answer, Key Homework 4 Dvid McIntyre Mr 25, So From Newton s Second Lw, (prt 2 of 2) 5 points Determine (m 1 + ) correct 2. None of these (m 1 + ) (m 1 + ) 3. he horse cn pull the buy forwrd only if the horse weihs more thn the buy. 4. he horse pulls before the buy hs time to rect so they move forwrd. Accordin to Newton s third lw, the force of the buy on the horse is s stron s the force of the horse on the buy. 010 (prt 1 of 1) 10 points he pulleys re frictionless nd mssless. he ccelertion of rvity is 9.8 m/s 2. he system is in equilibrium (m 1 + ) (m 1 + ) 1 2 m k (m 1 + ) Consider m 1 nd s whole object with mss (m 1 + ), then 11 k So (m 1 + ) 1 (m 1 + ) (m 1 + ). 37 k 009 (prt 1 of 1) 0 points Consider buy bein pulled by horse. Which is correct? Find the tension. Correct nswer: N. 1. he force on the buy is s stron s the force on the horse. he horse is joined to the Erth by flt hoofs, while the buy is free to roll on its round wheels. correct 2. he horse pulls forwrd slihtly hrder thn the buy pulls bckwrd on the horse, so they move forwrd. Let : m 1 37 k, 7.2 k, m 3 11 k. nd

4 Answer, Key Homework 4 Dvid McIntyre Mr 25, constnt speed, the scle reds U. When the elevtor is movin down with the sme constnt speed, the scle reds D. 3 m m 1 Scle he mss m 1 defines the tension 1 1 m 1. At the riht-hnd pulley, 1 cts down on either side of the pulley nd 2 cts up, so m 1. At the mss, 3 cts up, nd nd 2 ct down, so m 1. At the left-hnd pulley, 3 cts up on either side of the pulley nd 4 cts down, so At the mss m 3, m m 1. Choose the correct reltionship between the three scle redins. 1. U > W > D 2. U W D correct 3. U D, U > W 4. U < W < D 5. U D, U < W Consider the free body dirm for ech the cse where the elevtor is ccelertin down (left) nd up (riht). he mn is represented s sphere nd the scle redin is represented s. 4 + m 3 4 m 3 (2 + 4 m 1 m 3 ). m m 011 (prt 1 of 2) 0 points A mn stnds on scle in n elevtor. When the elevtor is t rest, the scle reds W. When the elevtor is movin upwrd with Note: he ccelertion in ll three cses is zero, so ccordin to Newton s Second Lw the force mesured by the scle is F scle W.

5 Answer, Key Homework 4 Dvid McIntyre Mr 25, (prt 2 of 2) 0 points Suppose the elevtor now ccelertes downwrd t constnt rte of Wht is the rtio of the new scle redin to the vlue W of the scle redin when the elevtor is t rest? Given : m k, k, µ 0. nd 1. F scle W F scle W F scle W F scle W 5. F scle W F scle W correct 7. F scle W 2.0 Newtons Second Lw in this cse reds W F scle 0.37 W, where W is the weiht of the mn. herefore, F scle W (prt 1 of 3) 4 points A block of mss k lies on frictionless tble, pulled by nother mss k under the influence of Erth s rvity. he ccelertion of rvity is 9.8 m/s k µ k Wht is the mnitude of the net externl force F ctin on the two msses? Correct nswer: N. N m 1 m 1 he net force on the system is simply the weiht of. F net ( k) (9.8 m/s 2 ) N. 014 (prt 2 of 3) 3 points Wht is the mnitude of the ccelertion of the two msses? Correct nswer: m/s 2. From Newton s second lw, F net (m 1 + ). Solvin for, m k k k (9.8 m/s2 ) m/s (prt 3 of 3) 3 points Wht is the mnitude of the tension of the rope between the two msses? Correct nswer: N. Anlyzin the horizontl forces on block m 1, we hve Fx : m 1 ( k) ( m/s 2 ) N.

6 Answer, Key Homework 4 Dvid McIntyre Mr 25, (prt 1 of 3) 4 points A 4.14 k block slides down smooth, frictionless plne hvin n inclintion of 35. he ccelertion of rvity is 9.8 m/s 2. lon tht xis. So, by Newton s second lw, Fx m m sin θ hus sin θ With our prticulr vlue of θ, 2.98 m 4.14 k µ 0 35 Find the ccelertion of the block. Correct nswer: m/s 2. (9.8 m/s 2 ) sin m/s (prt 2 of 3) 3 points Wht is the block s speed when, strtin from rest, it hs trveled distnce of 2.98 m lon the incline. Correct nswer: m/s. Since v 0 0, m/s 2 nd L 2.98 m, Given : m 4.14 k, θ 35, nd µ s 0. Consider the free body dirm for the block m sin θ N Bsic Concepts: µ N N m cos θ W m F x,net F cos θ W 0 W m sin θ m Motion hs constnt ccelertion. Recll the kinemtics of motion with constnt ccelertion. Solution: Becuse the block slides down lon the plne of the rmp, it seems loicl to choose the x-xis in this direction. hen the y-xis must emere perpendiculr to the rmp, s shown. Let us now exmine the forces in the x- direction. Only the weiht hs component v 2 f v (x x 0 ) 2 ( m/s 2 ) (2.98 m) v f m/s. It is very importnt to note t this point tht neither of these vlues depended on the mss of the block. his my seem odd t first, but recll wht Glileo discovered 300 yers o objects of differin mss fll t the sme rte. 018 (prt 3 of 3) 3 points Wht is the mnitude of the perpendiculr force tht the block exerts on the surfce of the plne t distnce of 2.98 m down the incline? Correct nswer: N. By exminin the free body dirm in, we see tht the force in the y direction is iven by F m cos θ FN. F N m cos θ 4.14 k (9.8 m/s 2 ) cos N.

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