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Transcript
PHY 131 Ch 1 to 5 A stone is dropped from a cliff. The graph that best represents its motion while it falls is: III A large cannon is fired from ground level over level ground at an angle of 30° above the horizontal. The muzzle speed is 980 m/s. The projectile will travel what horizontal dist before striking the ground? y = ½ a t2 + vo t + yo 0 = ½-10t2 + sin30(980) t + 0 t = 98 sec One object is thrown vertically upward with an initial velocity of 100 m/s and another object with an initial velocity of 10 m/s. The maximum height reached by the first object will be _______ that of the other. a = ∆v / ∆t -10 = 0 – 100 / t t = 10 sec a = ∆v / ∆t -10 = 0 – 10 / t t = 1 sec y = ½ a t 2 + vo t y = ½-10 t2 + 100t y = 500 m y = ½ a t 2 + vo t y = ½-10 t2 + 10t y=5m An object placed on an equal-arm balance requires 12 kg to balance it. When placed on a spring scale, the scale reads 12 kg. Everything is now transported to the Moon where free-fall acceleration is 1/6 that on Earth. The new readings of the balance and spring scale (respectively) are: -kx = mg (gmoon = 1.66 m/s2) Ans: 12 kg, 2 kg Two blocks weighing 250 N and 350 N respectively, are connected by a string that passes over a massless pulley as shown. The tension in the string is: FNet = mT a 350 – 250 = (25+35) a a = 1.7 m/s2 Accel down Accel UP FT = 35kg (10-1.7) OR 25kg (10+1.7) FT = 290 N An object is thrown straight down with an initial speed of 4 m/s from a window which is 8 m above the ground. The time it takes the object to reach the ground is: 2 y = ½ a t + vot + yo 0 = ½-10 t2 - 4t + 8 t = 0.93 sec A Ferris wheel with a radius of 8.0 m makes 1 revolution every 10 s. As he passed the top of the path he releases a ball. How far from the point on the ground directly under the release point does the ball land? h = ½ a t2 vx = ∆x / ∆t v = 2πr / T 2r = ½10 t2 5 = ∆x / 1.8 v = 2π8 / 10 t = 1.8 sec ∆x = 9 m bottom v = 5 m/s A ball is thrown horizontally from the top of a 20-m high hill. It strikes the ground at an angle of 45°. With what speed was it thrown? y = ½ a t2 20 = ½-10 t2 t = 2 sec a = ∆vy / ∆t -10 = (sin45v – 0)/2 v = 28 m/s Name: vx = ∆x /∆t cos30 (980) = ∆x /98 ∆x = 83000 m A ferry boat is sailing at 12 km/h 30° W of N with respect to a river that is flowing at 6.0 km/h E. As observed from the shore, the ferry boat is sailing: vx vy No x component River 6 0 Due North Ferry -sin30 12 cos30 12 Sum 0 10.4 When a 40-N force, parallel and directed up a ramp inclined by 30° above the horizontal, the acceleration of the crate is 2.0 m/s2, up the incline. The mass of the crate is: FNet = mT a 40 – sin30 mg = m (2) m = 5.7 kg A 25-kg crate is pushed across a frictionless horizontal floor with a force of 20 N, directed 20° below the horizontal. The accel of the crate is: FNet = mT a cos20 (20)= 25 a a = 0.75 m/s2 A 5-kg block is suspended by a rope from the ceiling of an elevator that accelerates downward at 3.0 m/s2. The tension force of the rope on the block is: FT = m (g + a) FT = 5 (10 + 3) FT = 65 N, up The position of an object as a function of time is given in meters by x = (at +bt2) i + (ct) j. What is its velocity as a function of time? v = dx / dt v = (a + 2bt) i + (c) j An object is thrown vertically into the air. Which of the following five graphs represents the velocity (v) of the object as a function of the time (t)? (Up is positive) III A projectile is shot vertically upward with a given initial velocity. It reaches a maximum height of 100 m. If on a second shot the initial velocity is doubled, then the projectile will reach a maximum height of: y=½ a t2 100 = ½-10 t2 t = √20 a = ∆v / ∆t -10 = 0 – 90 / t t = 9 sec a = ∆v / ∆t -10 = 0 – vo / √20 vo = 45 m/s y = ½ a t 2 + vo t y = ½-10 t2 + 90t y = 400 m When a 40-N force, parallel and directed up a ramp inclined by 30° above the horizontal, the acceleration of the crate is 2.0 m/s2, down the incline. The mass of the crate is: FNet = mT a 40 – sin30 mg = m (-2) m = 13.8 kg v2 = vx2 + vy2 v2 = (14-6)2 + 62 v = 10 mph A boat is traveling upstream at 14 mph with respect to a river that is flowing at 6 mph. A man runs directly across the boat, from one side to the other, at 6 mph. The speed of the man with respect to the ground is: A girl jogs around a horizontal circle with a constant speed. She travels one fourth of a revolution, a distance of 25 m along the circumference of the circle, in 5.0 s. The magnitude of her acceleration is: ¼(2πr) = 25m a = v2 / r r = 50/π a = (25/5)2 / 15.9 r = 15.9 m a = 1.6 m/s2 A dart is thrown horizontally at 20 m/s. It hits 0.1 s later. What is the distance the dart drops during its path? y = ½ a t2 y = ½10 .12 y = 0.05 m Two forces, one with a magnitude of 3 N and the other with a magnitude of 5 N, are applied to an object. For which orientation of the shown forces shown is the magnitude of the acceleration of the object the least? We definitely know: IV is Greatest, and I is the Least A baseball is hit straight up and is caught by the catcher 2.0 s later, at the same height at which it left the bat. The maximum height of the ball during this interval is: ttop2 2 y=½a y = ½10 1 y=5m A crate rests on a horizontal surface and a woman pulls on it with a 10-N force. Rank the situations shown below according to the magnitude of the normal force exerted by the surface on the crate, least to greatest. 3, 2, 1 A car drives in a straight line on a level road with a constant acceleration of 3 m/s2. A ball is suspended by a string from the ceiling of the car; the ball does not swing, being at rest with respect to the car. What angle does the string make with the vertical? tan θ = 3/10 θ = 17° A 90-kg man stands in an elevator that is moving up at a constant speed of 5.0 m/s. The force exerted by him on the floor is about: FW = m (g + a) FW = 90 (10 + 0) FW = 900N A massless rope passes over a massless pulley suspended from the ceiling. A 4-kg block is attached to one end and a 5-kg block is attached to the other end. The acceleration of the 5-kg block is: FNet = mT a 5g – 4g = (4+5) a a = g/9 m/s2 An object is dropped from rest. Which of the five following graphs correctly represents its motion? The positive direction is taken to be upward. Take careful note of the axes. An object is thrown vertically upward with a certain initial velocity in a world where the acceleration due to gravity is 19.6 m/s2. The height to which it rises is ____ that to which the object would rise if thrown upward with the same initial velocity on the Earth. half a = ∆v / ∆t -10 = 0 – 100 / t t = 10 sec y = ½ a t2 + vo t y = ½-10 t2 + 100t y = 500 m a = ∆v / ∆t -20 = 0 – 100 / t t = 5 sec y = ½ a t2 + vo t y = ½-20 t2 + 100t y = 250 m A short 10-g string is used to pull a 50-g toy across a frictionless horizontal surface. If a 0.030 N force is applied horizontally to the free end of the string, the force of the string on the toy, at the other end, is: Newton’s 3rd law: 0.030 N A motor boat can travel at 10 km/h in still water. A river flows at 5 km/h west. A boater wishes to cross from the south bank to a point directly opposite on the north bank. At what angle must the boat be headed? tan θ = 5 / √75 vx vy θ = 30° River 5 0 Boat -5 102 = -52 + vy2 Sum 0 vy = √75 A stone is thrown vertically upward with an initial speed of 19.5 m/s. It will rise to a maximum height of: A stone is tied to a 0.50-m string and whirled at a constant speed of 4.0 m/s in a vertical circle. The acceleration at the bottom of the circle is: a = v2/r = 32m/s2, up FT – mg = mv2/r FT = 5mg a = ∆v / ∆t -10 = 0 – 19.5 / t t = 1.95 sec A boy on the edge of a vertical cliff 20 m high throws a stone horizontally outwards with a speed of 20 m/s. It strikes the ground at what horizontal distance from the foot of the cliff? y=½ a t2 20 = ½10 t2 t = 2 sec vx = ∆x / ∆t 20 = xo – 0 / 2 xo = 40 m A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be: F= ma a = ∆v /∆t 8 = 16 a ½ = vf – 0 / 4 a = ½ m/s2 vf = 2 m/s y = ½ a t2 + vo t y = ½-10 t2 + 19.5t y = 19 m A 100 gram pendulum bob is held at an angle θ from the vertical by a 2 N horizontal force F. The tension in the string is: tan θ = 2/1 θ = 63° FT2 = 22 + 12 FT = √5 Two forces are applied to a 5.0-kg crate; one is 6.0 N to the north and the other is 8.0 N to the west. The magnitude of the acceleration of the crate is: A particle goes from x = –2 m, y = 3 m, z = 1 m to x = 3 m, y = –1 m, z = 4 m. Its displacement is: a2 = 6/52 + 8/52 a = 2 m/s2 (3 - -2), (-1 – 3), (4 - 1) (5, -4, 3) A 32-N force, parallel to the incline, is required to push a certain crate at constant velocity up a frictionless incline that is 30° above the horizontal. The mass of the crate is: FNet = mT a 32 – sin30 mg = m (0) m = 6.4 kg (45%) Below is a ramp inclined by 30.0 degrees, initially at rest. A Force, F = 30.0 N, is applied by a rocket (unlimited fuel). The 2.00 block will travel another 3 meters up the ramp before going over the edge. The coordinates of the edge are (0, 0), up + and right +. (a) What is the normal 2% 2% 8% force exerted by the FN = cos30(20) + F sin30 ramp on the block? FN = 32 Newtons (b) If a landing pad is 1 meter below the edge, how far will the block travel in the horizontal direction? 8% FNet 3% = mT a 2% 1% Fcos30 – sin30 mg = m a 26 - 10 = 2 (a) a = 8 m/s2 x = ½ a t2 2% 3 = ½ 8 t2 t = 0.866 sec On Ramp Free Fall with Rocket a = ∆v / ∆t 8 = vf – 0 / .866 vf = 6.93 m/s 2% vx = cos30(6.93) vx = 6 m/s 1% Fx = m ax 30 = 2 (ax) ax = 15 m/s2 vy = sin30(6.93) vy = 3.46 m/s 1% 5% y = ½ a t 2 + v o t + yo 1% 1% 1% 1% 0 = ½-10t2 + 3.46 t + 1 t = 0.91 sec x = ½ a t2 + vo t + yo 1% 1% 1% 1% x = ½ 15 .912 + 6(.91) + 0 x = 13.1 meters F = 20 N (a) 2% 2% On Ramp 8% FN = 20N + sin30 F FN = 30 Newtons FNet 6% x = ½ a t2 3 = ½ 8.66 t2 t = 0.832 sec = mT a 1% a = ∆v / ∆t 8.66 = vf – 0 / .832 vf = 7.21 m/s 2% Fcos30 = m a 17.3 = 2 (a) a = 8.66 m/s2 F = 8.00 N (a) 2% 2% 8% FN = 20cos30 + sin30 F FN = 21.6 Newtons (b) 2% On Ramp FNet 3% 8% = mT a 1% mg sin30 - Fcos30 = m a 3.07 = 2 (a) 2 a = 1.5 m/s x = ½ a t2 3 = ½ 1.54 t2 t = 2.0 sec 2% 2% Free Fall Fx = m ax cos30 F = 2 (ax) ax = 8.66 m/s2 5% y = ½ at2 + vot + yo 1% 1% 1% 0 = ½-15t2 + 0 + 1 t = 0.365 sec Fy = m ax 7% 1% 1% sin30F + mg = 2 (ay) ay = -15 m/s2 End of Ramp Free Fall a = ∆v / ∆t 1.5 = vf – 0 / 2.0 vf = 3.0 m/s 2% Fx = m ax -8 = 2 (ax) ax = -4 m/s2 vx = cos30(3) vx = 2.6 m/s y = ½ a t 2 + v o t + yo 1% 1% 1% 1% 0 = ½10t2 + 1.5 t - 1 t = 0.321 sec vy = sin30(3) vy = 1.5 m/s 1% 1% 5% x = ½ a t2 + vo t + yo 1% 1% 1% 1% x = ½ -4 .3212 + 2.6(.321) + 0 x = 0.64 meters x = ½ a t2 + vo t + yo 2% 1% 1% 1% x = ½ 8.66 .3652 + 7.21(.365) + 0 x = 3.21 meters