PY1052 Problem Set 3 Autumn 2004 Solutions

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1 PY1052 Poblem Set 3 Autumn 2004 Solutions C F = 8 N F = 25 N 1 2 A A (1) A foce F 1 = 8 N is exeted hoizontally on block A, which has a mass of 4.5 kg. The coefficient of static fiction between A and the table is µ s = 0.20, and the coesponding coefficient of kinetic fiction is µ k = (a) Does the block move? What is the fiction foce felt by the block? Now a foce F 2 = 25 N is exeted hoizontally on block A. (b) What is the minimum mass of block C to keep A fom sliding? (c) Block C is suddenly lifted off. What is the acceleation of block A? (a) The maximum possible foce of static fiction between A and the table is F s,max = µ s N = µ s M A g = (0.20)(4.5 kg)(9.8 m/s 2 ) = 8.82 N. This means that if a foce is applied hoizontally on block A, the block will move if this foce is at least 8.82 N, and will not move if this foce is less than 8.82 N. Theefoe, the block does not move due to the action of F 1. Since the block is not moving, the sum of foces in the hoizontal diection must be zeo, and so the foce of static fiction is exactly equal and opposite to F 1 : 8.0 N to the left. (b) The maximum possible foce of static fiction between A and the table when block C is sitting on block A is F s,max = µ s N = µ s (M A + M C )g. In ode fo the two blocks not to move unde the action of F 2, the mass of C must be lage enough that F 2 is smalle than the maximum possible foce of static fiction. The coesponding minimum mass fo C is given by F 2 µ s (M A + M C )g = 0 M C = F 2 µ s g M A = 25 N 0.20(9.8 m/s kg ) M C = 8.25 kg (c) The acceleation of A afte C is lifted off will be given by F 2 µ k M A g = M A a a = F 2 M A µ k g = 25 N 4.5 kg 0.15(9.8 m/s2 ) a = 4.08 m/s 2

2 (2) A 110-gam hockey puck sent sliding ove ice with an initial speed of 6.0 m/s is stopped in 15 m by the fictional foce on it fom the ice. (a) What is the magnitude of the fictional foce on the ice? (b) What is the coefficient of kinetic fiction between the puck and the ice? (a) Since the fictional foce is constant, the acceleation will be constant, and we can use ou equations fo constant acceleation. We know the initial velocity (6.0 m/s), the final velocity (0 m/s) and the distance tavelled (15 m). Theefoe, v 2 = v 2 o + 2a(x x o ) a = v2 v 2 o 2(x x o) a = m/s 2 2(15 m) = 1.2 m/s 2 The magnitude of the fictional foce will be F k = m a = (0.110 kg)(1.2 m/s 2 ) = N (b) The fictional foce is given by F = µ k N In the vetical diection, thee ae two foces, the puck s weight and the nomal foce, which must add to zeo: N mg = 0 N = mg Theefoe, F = µ k N = µ k mg µ k = F mg = ma mg = a g = 1.2 m/s2 9.8 m/s 2 = 0.12 (3) A pig slides down a 35 incline in twice the time that it would take him to slide down a fictionless 35 incline. What is the coefficient of kinetic fiction between the pig and the incline? Thee ae thee foces acting on the pig: the nomal foce, its weight and the fiction foce. The nomal foce N is puely pependicula to the amp, the fictional foce F k is puely up the amp and the weight has a component mg sin(35) down the amp and mg cos(35) pependicula to the amp, opposite the nomal foce. Thee is no acceleation pependicula to the amp, and we ll call the acceleation down the amp a f (fo acceleation with fiction): N mg cos(35) = 0 pependicula to the amp mg sin(35) F k = ma f along the amp mg sin(35) µ k N = ma f Solving fo N in the fist equation, plugging into the thid equation and solving fo a f yields a f = g(sin(35) µ k cos(35)) If thee is no fiction, this is equivalent to µ k = 0, and the coesponding acceleation a nf will be a nf = g sin(35)

3 Now use these two acceleations to find the time fo the pig to slide down the amp of length x with and without fiction, t f and t nf : x = v 0 t at2 = 1 2 at2 v 0 = 0 x = 1 2 a ft 2 f = 1 2 a nft 2 nf But we ae told that t f = 2t nf : 1 2 a f[2t nf ] 2 = 1 2 a nft 2 nf 4a f = a nf 4g(sin(35) µ k cos(35)) = g sin(35) 4µ k g cos(35) = 3g sin(35) µ k = 3 tan(35) = (4) A dat is thown hoizontally with an initial speed of 10 m/s towad the bull s eye on a dat boad. It hits at a point diectly below the bull s eye 0.19 s late. (a) How fa below the bull s eye does the dat hit? (b) How fa away fom the dat boad is the dat eleased? (c) What ae the hoizontal and vetical components of the dat s velocity when it hits the boad? We have the following two sets of equations fo the dat s motion in the x and y diections (we take the initial position of the dak to be x = 0, y = 0, and will take +y to be upwad and +x to be in the diection of motion of the dat): x = x 0 + v 0x t x = 10 m/st v x = v 0x v x = 10 m/s y = y 0 + v 0y t 1 2 gt2 y = 1 2 gt2 v y = v 0y gt v y = gt (a) Because we know the time the dat is in flight, we can find the y value coesponding to this time: y = 1 2 gt2 = 1 2 (9.8 m/s2 )(0.19 s) 2 y = m = 17.7 cm The dat hits 17.7 cm below the bull s eye. (b) Knowing the time of flight, we can also find the total distance to the boad: x = 10 m/st = 10 m/s(0.19 s) = 1.9 m (c) We can also find the hoizontal and vetical components of the dat s velocity when it hits the boad: v x = 10 m/s v y = gt = (9.8 m/s 2 )(0.19 s) = 1.86 m/s

4 (5) A supply aiplane diving at an angle of 35.0 with the hoizontal eleases a package of supplies at an altitude of 130 m. The package hits the gound 3.5 s afte being eleased. (a) What is the speed of the plane? (b) How fa did the package tavel hoizontally duing its flight? We have the following two sets of equations fo the package s motion in the x and y diections (we take the initial position of the package when it is eleased to be x = 0, y = 0, and will take +y to be upwad and +x to be in the hoizontal diection of motion of the plane): x = x 0 + v 0x t x = v cos(35) m/st v x = v 0x v x = v cos(35) m/s y = y 0 + v 0y t 1 2 gt2 y = v sin(35)t 1 2 gt2 v y = v 0y gt v y = v sin(35) gt (a) We know that y = 130 m when t = 3.5 s. We wish to find v we can do this using the equation fo y: y = v sin(35)t 1 2 gt2 y f = v sin(35)t f 1 2 gt2 f v sin(35)t f = y f 1 2 gt2 f v = y f 1 2 gt2 f sin(35)t f = ( 130 m) 1 2 (9.8 m/s2 )(3.5 s) 2 sin(35)(3.5 s) v = 34.8 m/s (b) Knowing v and t f, we can also find the hoizontal distance tavelled befoe the package hit the gound: x f = v cos(35) m/st = (34.8 m/s) cos(35)(3.5 s) x f = 99.8 m (6) The Sun has a adius of m and the mateial at its equato otates about its axis once evey 26 days. What is the linea velocity of mateial at the Sun s equato due to its otation? The linea velocity will be the equal to v = 2πR T = 2π( m) 26 d We must simply expess the peiod T in seconds to get the speed in m/s: T = 26 d 24 hs d v = 2π( m) s 3600 s h = s = m/s = 1.94 km/s

5 (7) On a Fench TGV tain, the magnitude of the acceleation expeienced by the passenges is to be limited to 0.050g. (a) If such a tain is going aound a cuve at a speed of 216 km/h what is the smallest adius of cuvatue that the cuve can have without exceeding the maximum allowed acceleation on the passenges? (b) With what speed does the tain go aound a cuve with a 1.00 km adius if the acceleation exeted on the passenges is at its maximum allowed value? (a) Since the tain is moving in a cicula ac, we have a c = v2 v2 = a c We can see that the maximum centipetal acceleation, 0.050g, coesponds to the minimum adius. The tain s speed and acceleation ae v = 216 km 1000 m 1 h = 60.0 m/s h 1 km 3600 s a c = 0.050g = 0.050(9.8 m/s 2 ) = m/s 2 Theefoe, the minimum adius is (60.0 m/s) m/s 2 = v2 a c = = m = 7.35 km (b) Again we have a c = v2 v = a c v = ( m)(0.490 m/s 2 ) = 22.1 m/s m M (8) A puck of mass m slides in a cicle on a fictionless table while attached to a hanging mass M by a cod though a hole in the table. With what speed must the puck move to keep the hanging mass stationay? The upwad tension in the cod must balance the downwad foce of gavity acting on M. This tension will also be equal to the centipetal foce acting on m: T Mg = 0 balance of foces fo M T = mv2 centipetal foce on m mv 2 Mg = 0 v = M m g

6 (1b) Fo women s volleyball, the top of the net is 2.24 m above the floo and the cout measues 9.0 m by 9.0 m on each side of the net. Using a jump seve, a playe stikes the ball at a point that is 3.0 m above the floo and a hoizontal distance of 8.0 m fom the net. If the initial velocity of the ball is hoizontal, what minimum magnitude must it have if the ball is to clea the net, and what magnitude can it have if the ball is to stike the floo inside the back line on the othe side of the net? 3.0 m 2.24 m 8.0 m 9.0 m Take the point whee the volleyball is hit to be (x = 0, y = 3.0 m). We also know that v 0,x = v 0 and v 0,y = 0. Ou equations fo x and y as a function of time ae then x = v 0 t constantvelocity y = 3.0 m 1 2 gt2 constantacceleation In ode to clea the net, we must have y = 2.24 m when x = 8.0 m. Use the elation fo y to find the time, then plug this into the elation fo x to find v 0. t = 2.24 m 3.0 m g/ m = 3.0 m 1 2 gt2 = s v 0 = x t = 8.0 m = 20.3 m/s s Thus, the minimum speed the volleyball must have is 20.3 m/s. Fo the ball to land on the line, we must have x = 17.0 m when y = 0. Using the same pocedue, we find t = 0 m 3.0 m g/2 = s 0 m = 3.0 m 1 2 gt2 v 0 = x t = 17.0 m = 21.7 m/s s Thus, the seve must hit the volleyball so that it has a speed no less than 20.3 m/s and no moe than 21.7 m/s. (2b) A gadene wishes to pile a cone of sand onto a cicula aea in his yad. The adius of the cicle is R, and no sand is to spill onto the suounding aea. If µ s is the coefficient of static fiction between each laye of sand along the slope and the sand beneath it (along which it must slip), show that the geatest volume of sand that can be stoed in this manne is πµ s R 3 /3. The volume of a cone is Ah/3, whee A is the base aea and h is the cone s height.

7 (3b) A 1000-kg boat is taveling at 90 km/h when its engine is shut off. The magnitude of the fictional foce between the boat and wate is popotional to the speed of the boat: f k = 70v, with v in m/s and the foce in Newtons. How long does it take fo the boat to slow to 45 km/h? We can find a elation between the time and the change in the velocity as follows: F = kv = ma = m dv dt dt = m dv k v Integating both sides of this equation, we obtain tf t i dt = m k vf v i t f t i = m k ln v f v i dv v = m k ln v i v f Inseting m = 1000 kg, k = 70 Ns/m, v i = 90 km/h and v f = 45 km/h, we find t = ln 45 = 1000 ln 2 = 9.90 s 70

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