PHYSICS 111 HOMEWORK SOLUTION #5. March 3, 2013


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1 PHYSICS 111 HOMEWORK SOLUTION #5 Mach 3, 2013
2 0.1 You 3.80kg physics book is placed next to you on the hoizontal seat of you ca. The coefficient of static fiction between the book and the seat is 0.650, and the coefficient of kinetic fiction is You ae taveling fowad at 72.0 km/h and bake to a stop with constant acceleation ove a distance of 30.0 m. You physics book emains on the seat athe than sliding fowad onto the floo. Is this situation possible? a) Lets s look at the foces exeted on the physics book: +y N f fic +x m 1 g The acceleation of the ca can be calculated using v 2 f v2 i = 2a x v 2 a = 2 x = ( ) = 6.67m/s 2 On the othe hand, Pojecting 2nd Law on the yaxis gives N=mg; Fo the book to slide off the seat, acceleation should ovecome fction: f s < ma µ s N < ma µ s mg < ma µ s = < a g = = 0.68 This is valid and the book will definitely slide fowad to the floo. 2
3 A 2.70kg block stats fom est at the top of a 30.0 incline and slides a distance of 1.90 m down the incline in 2.00 s. a) Find the magnitude of the acceleation of the block. b) Find the coefficient of kinetic fiction between block and plane. c) Find the fiction foce acting on the block. d) Find the speed of the block afte it has slid 1.90 m. a) +y N fk +x α α m g We can use x = 1 2 at2 + v 0 t + x 0 to find the acceleation of the object as it slides down. with v 0 = and x 0 = 0. The object slides 1.90 metes in 2 seconds, this should give us: b) a = 2x t 2 = = 0.95m/s 2 Newton Second Law: Fi = m g + F fic = m a Pojection on the xaxis : mg sin α f k = ma (*) Pojection on the yaxis : mg cos α N = 0 (**) 3
4 We bea in mind that f k and N ae connected : f k = µ k N Fom (*) and (**) we get : µ k = f k N = mg sin α ma mg cos α = tan α a g cos α 0.95 = tan(30) 9.81 cos(30) = c) The fiction foce acting on the block: f k = µ k N = µ k mg cos α = cos(30) = 10.7N d) Speed of the block afte sliding 1.90m We can use the time independent equation : v 2 v 2 0 = 2a x v = v = 1.9m/s 4
5 A 9.80kg hanging object is connected by a light, inextensible cod ove a light, fictionless pulley to a 5.00kg block that is sliding on a flat table. Taking the coefficient of kinetic fiction as 0.185, find the tension in the sting. +y N T 2 f k T1 +x m 1 g m 2 g + T 1 = T 2 = T ; a 1 = a 2 = a and f k = µ k N Pojecting Newton s second law fo object m 1 gives us f k + T = m 1 a and N m 1 g = 0 Pojecting Newtons s Law fo object m 2 on the yaxis gives us m 2 g T = m 2 a theeby, a = m2g T m 2 5
6 T = m 1 a + f k = m 1 a + µ k N = m 1 g m 1 T + µ k m 1 g m 2 m 1 + m 2 T m 2 = gm 1 (1 + µ k ) Finally, T = m 1m 2 g(1 + µ k ) m 1 + m ( ) = = 38.48N 0.4 Two blocks connected by a ope of negligible mass ae being dagged by a hoizontal foce (see figue below). Suppose F = 73.0 N, m1 = 14.0 kg, m2 = 26.0 kg, and the coefficient of kinetic fiction between each block and the suface is a) Daw a feebody diagam fo each block. b) Detemine the acceleation of the system. c) Detemine the tension T in the ope. 6
7 0.4. a)fee body diagam fo each block b) Pojecting Newton s 2nd Law of the two blocks on the yaxis gives: n 1 = m 1 g and n 2 = m 2 g On the xaxis fo m 1 : T f 1 = m 1 a; with f 1 = µ k n 1 we get T = µ k n 1 + m 1 a = µ k n 1 g + m 1 a On the xaxis fo m 2 and with f 2 = µ k n 2 m 2 a = F T f 2 = F (f 1 + m 1 a) f 2 = F µ k n 1 m 1 a µ k n 2 Beaing in mind that n 1 = m 1 g and n 2 = m 2 g we should finally get: a = F µ kg(m 1 + m 2 ) m 1 + m ( ) = = 0.94m/s 2 7
8 c) Tension in the ope T = µ k m 1 g + m 1 a = m 1 (µ k g + a) = 14( ) = 25.52N 0.5 A block of mass 1.75 kg is pushed up against a wall by a foce P that makes an angle of θ = 50.0 angle with the hoizontal as shown below. The coefficient of static fiction between the block and the wall is a) Detemine the possible values fo the magnitude of P that allow the block to emain stationay. (If thee is no maximum, ente NONE in that answe blank.) b)what happens if P has a lage value than P max? c) )What happens if P has a smalle value than P min? d) Repeat pats (a) and (b) assuming the foce makes an angle of θ = 12.2 with the hoizontal. 8
9 0.5. a) We will dwas two fee body diagams fo this poblem, and the eason is that the fiction foce will eithe push the object up if it slides down o it will push the object down as it slides up. P P f s N θ N θ f s mg mg Fo the object to be stationay in the sitution at left, 2nd Law equies: mg + f s = p sin θ and with f s = µ s N N = p cos θ mg + µ s N cos θ = p sin θ p(sin θ µ s cos θ) = mg p = mg sin θ µ s cos θ p = sin cos 50 P max = 28.67N Fo the object to be stationay in the sitution at ight, 2nd Law equies: mg f s = p sin θ and with f s = µ s N N = p cos θ mg µ s N cos θ = p sin θ 9
10 b) p(sin θ + µ s cos θ) = mg p = mg sin θ + µ s cos θ p = sin cos 50 P min = 18.60N If p has a lage value than p max the object will slide up the wall. c) If p has a smalle value than p min the object will slide down the wall. d) If the angle changes to 12.2 we will have: p max = sin cos 12.2 = negative value!!!!! The block will not slide along the wall and p min = sin cos 12.2 = 36.88N The block is capable of sliding down the wall unde this value of P min 0.6 A cuve in a oad foms pat of a hoizontal cicle. As a ca goes aound it at constant speed 14.0 m/s, the hoizontal total foce on the dive has magnitude 149 N. What is the total hoizontal foce on the dive if the speed on the same cuve is 23.9 m/s instead? Let s call F 1 and F 2 the foces exeted at speed v 1 and speed v 2 espectively. The two ae centipetal foces of magnitudes mv2. we have: F 2 = mv2 2/ F 1 mv1 2/ F 2 = v2 2 F 1 v1 2 = 149( )2 10
11 0.6. = N 11
12 0.7 Conside the following figue. Why is the following situation impossible? The object of mass m = 4.00 kg in the figue above is attached to a vetical od by two stings of length = 2.00 m. The stings ae attached to the od at points a distance d = 3.00 m apat. The object otates in a hoizontal cicle at a constant speed of v = 3.00 m/s, and the stings emain taut. The od otates along with the object so that the stings do not wap onto the od. T d a c α α T mg 12
13 0.7. a) A pojection of Newton s 2nd Law on the centipetal diection gives: T 1 cos α + T 2 cos α = ma c = mv2 A pojection on the vetical diection gives: T 1 sin α T 2 sin α mg = 0 The geomety of the poblem equies = l 2 ( d 2 )2 = = 1.32m cos α = l = = 0.66 sin α = d/2 l = = 0.75 Back to ou fits equations we solve fo T 1 and T T T 2 = mv2 = = T T 2 = mg = = We should get T 1 = 46.81N and T 2 = 5.5N!!! This value of T 2 is not possible in ou situation and it implies that the lowe sting is pushing the object athe than pulling it. 13
14 b) The vaiables in this poblem that can make this situation physically acceptable ae the speed of the object and the g value of gavity. In Mas fo instance g = 3.711m/s 2, ou eqations become 0.66T T 2 = mv2 = = T T 2 = mg = = This gives T 1 = 30.55N and T 2 = 10.76N. 14
15 A coin placed 30.2 cm fom the cente of a otating, hoizontal tuntable slips when its speed is 50.2 cm/s. a) What foce causes the centipetal acceleation when the coin is stationay elative to the tuntable? b) What is the coefficient of static fiction between coin and tuntable? a) N f s a c mg a) The centipetal foce is povided by the static fiction f s 15
16 b) Pojecting Newton s 2nd Law on the vetical gives mg = N On the centipatal diection : f s = µ s N = m v2 µ s mg = m v2 µ s = v2 g = ( ) = A hawk flies in a hoizontal ac of adius 14.2 m at a constant speed 4.10 m/s. a) Find its centipetal acceleation. b) It continues to fly along the same hoizontal ac, but inceases its speed at the ate of 1.00 m/s2. Find the acceleation in this situation at the moment the hawk s speed is 4.10 m/s. a) The centipetal acceleation is : a c = v2 = = 1.18m/s 2 16
17 0.10. b) a c a t The hawk is now stating to acceleate along the hoizontal with a tangential acceleation a t = 1.00m/s 2. The oveall vecto aceleation of the hawk the moment its speed was 4.10m/s becomes a = a c + a t having a magniutde of: a = a 2 c + a 2 t = = 1.54m/s A 43.0kg child swings in a swing suppoted by two chains, each 2.98 m long. The tension in each chain at the lowest point is a)find the child s speed at the lowest point. b)find the foce exeted by the seat on the child at the lowest point. (Ignoe the mass of the seat.) a) At some time of this motion the child on his seat will be on a vetical position, this will help us deive equations easily: The seat and the child as one object ae moving with same speed. 17
18 T T a c = mv2 mg We should have : T + T mg = mv2 and 2T v = m g = = 4.42N b) The child is now ou sole object, expeiencing only gavity mg and the nomal foce N exeted by the seat: we should have: N mg = mv2 N = mg + mv2 = 2T 704N 18
19 A ollecoaste ca has a mass of 499 kg when fully loaded with passenges. The path of the coaste fom its initial point shown in the figue to point B involves only upanddown motion (as seen by the ides), with no motion to the left o ight. Assume the ollecoaste tacks at points A and b ae pats of vetical cicles of adius 1 = 10.0 m and 2 = 15.0 m,espectively. a) If the vehicle has a speed of 19.1 m/s at point A, what is the foce exeted by the tack on the ca at this point? b) What is the maximum speed the vehicle can have at B and still emain on the tack? a) On point A, the foce exetad by the tack is the nomal which is also the centipetal foce and we have: N mg = mv2 1 N = m(g + v2 1 ) = 499( = 23099N 19
20 b On B, the nomal foce pointing downwas along with gavity should go to zeo to allow the ollecoaste to emain in tack, gavity is thus the only centipetal foce: mg = mv2 max 2 v max = 2 g v max = = 12.13m/s 20
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