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PHY131 Ch 6 to 10 Exam Name: Answer 3 of the below 4; clearly mark out the one your skip. One of the below can receive bonus points . 6.17 10.16 A massless spring hangs from the ceiling. A You push downward on a trunk at an angle 25° below 10 kg bike hangs from the spring. How much does its the horizontal with a force of 750 N. If the trunk is potential energy increase, if another 20 kg bike is also on a flat surface and the coefficient of static friction attached to it? The spring constant is 4000 N/m. is 0.66, what is the most massive trunk you will be able FW = mg mg = kx0 mg = kxf to move? FHookes = -kx 100 = 4000 x0 300 = 4000 x0 84 kg 73 kg 66 kg 101 kg x0 = 1/40 m x0 =3/40 m FPush-x - Ff = mT a (just over 0 to move) ∆Ug = ½ k ( xf2 - x 02 ) cos25°(750)) - 0.66 (mg + sin25°(750)) = 0 2 ∆Ug = ½(4000) ((3/40) – (1/40)2) m = 72.7 kg ∆Ug = 10 Joules 10.19 A block starts from rest at the top of a 40° 9.17 inclined plane and encounters a spring, of constant 2.4 A car heading north collides at an intersection with a kN/m, rigidly attached to the plane. If the block's truck of equal mass heading east. If they lock mass is 34 kg and it compresses the spring by 32 cm, together and travel at 28 m/s at 35° north of east find the distance the block traveled before it just after the collision, how fast was the car initially encountered the spring. traveling? 16 m/s 64 m/s 24 m/s 32 m/s m g h = ½kx2 m g (0.32+d)sin40° = ½kx2 m vy + 0 = (2m) sin 35° 28m/s d = 24.2 vy = 32.1 m/s Spring compression is the key to this problem, this is not a stiff spring, and must include in problem. Ch7 What is the acceleration of the system? (6(g) – 8(g) sin30°) – Ff = mT a (μ is equal on both surfaces) (60 – 40) – [0.1 (6(g) + 8(g)cos30°] = (6+6+8) a a = 0.35 m/s2 9.3 A stationary 2.68 kg object is struck by a stick. The object F dt = ∆mv experiences a horizontal force given by F = at - bt3 , where t is the time in 1500∫t dt – 20∫t3 dt = ∆mv milliseconds from the instant the stick first contacts the object. If a = 1500 ∆v = 1.996 m/s N/ms and b = 20 N/ms3, what is the speed of the object just after it comes away from the stick (t = 2.74 ms)? 2.3 m/s 16 m/s 2 m/s 14 m/s 8.6 The figure shows two wires that are tied to a 1000 g mass which revolves in a horizontal circle at a constant speed of 8.0 m/s. What is the tension in the upper wire? FT-up-y = FT-down-y + m g sin30° FT-up = cos30°FT-down + 10 N FT-up = FT-down + 20 N FT-up-x + FT-down-x = mv2 / r cos30° FT-up + cos30°FT-down = 1 82 / 0.866 FT-down = 32.7 N FT-up = 52.7 N PHY131 Ch 6 to 10 Exam Name: Answer 3 of the below 4; clearly mark out the one your skip. One of the below can receive bonus points . 6.12 10.41 Kieran takes off down a 50 m high, 10° slope on his A 50.0 gram ice cube can slide without friction up and jet-powered skis. The skis have a thrust of 410 N. down a 30.0 degree slope. The ice cube is pressed The combined mass of skis and Kieran is 50 kg (the against a spring at the bottom of the slope, fuel mass is negligible). Kieran's speed at the bottom compressing the spring 10.0 cm. The spring constant is is 40 m/s. What is the coefficient of kinetic friction 20.0 N/m. When the ice cube is released, what of his skis on snow? distance will it travel up the slope before reversing 0.74 0.87 0.90 0.67 direction? Wspring = Δ m g h d = 50/sin10° ½ k x2 = m g (sin30)(d + 0.1) ½(20)(0.1)2 = .05(10)(sin30)(d + 0.1) mg h + F d – Ff d = ½ m v2 d = 0.3 m m(g) 50 + 410(d) - μ(cos10°)mg(d) = ½ m 402 μ = 0.734 Spring compression is the key to this problem. 10cm is significant. 10.22 10.3 A fuel tank explodes, ripping the container into two A prankster drops a water balloon from the top of a pieces of masses 16 kg and 30 kg and sending them building on an unsuspecting person on the sidewalk 10.8 m and 7.5 m high, respectively. What was the below. If the balloon is traveling at 31.5 m/s when it energy released by the explosion if all energy becomes strikes a person's head (1.5 m above the ground), how the KE of the fragments? tall is the building? Neglect air resistance. 400 kJ 400 J 3.9 kJ 3900 kJ 54 m 52 m 51 m 55 m ΔK = m1 g h1 + m2 g h2 m g (h – 1.5) = ½ m (31.5)2 ΔK = 16(10)10.8 + 30(10)7.5 = h = 52.1 meters ΔK = 3980 Joules Ch7 What is the acceleration of the (8(g) sin30° - 4g sin60°) – Ff = mT a system? (μ is equal on both surfaces) ( 40 – 34.6) – [0.1 (8(g)cos30° +4(g)cos60°)] = (8+4) a 5.4 - 8.9 = 12 a a = 0 m/s2 9.3 A stationary 2.14 kg object is struck by a stick. The F dt = ∆mv object experiences a horizontal force given by F = at - bt3 1500∫t dt – 20∫t3 dt = ∆mv , where t is the time in milliseconds from the instant the (a) 17 m/s ∆v = 2.500 m/s stick first contacts the object. If a = 1500 N/ms and b = (b) 2.9 m/s 20 N/ms3, what is the speed of the object just after it (c) 20 m/s comes away from the stick (t = 2.74 ms)? (d) 2.5 m/s 8.4 A 100 g bead on a 60 cm long string is swung in a vertical circle about a point 200 cm above the floor. The tension in the string when the bead is at the very bottom of the circle is 6.0 N. A very sharp knife is suddenly inserted, as shown in the figure, to cut the string directly below the point of support. How far to the right of the center of the circle does the ball hit the floor? 2 / 2 3/3 8 / 8 8 / 8 4/4 2 2 d =½gt Fc = FT – mg = m v / r vx = Δx / Δt (2 – 0.6) = 5 t2 6 – 1 = 0.1 (vx)2 / 0.6 5.48 = Δx / 0.53 t = 0.53 sec vx = 5.48 m/s Δx = 2.90 meters