ExamName___________________________________MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.Solve the problem. (Use g = 9.8 m/s2.)1)A 21 kg box must be slid across the floor. If the coefficient of static friction between the boxand floor is 0.37, what is the minimum force needed to start the box moving from rest?A)53 NB)106 NC)76 ND)205 N1)Formula: F = coefficent of friction*m*g2)A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. Ifthe coefficient of kinetic friction is 0.19, at what rate does the box accelerate down the slope?A)5.5 m/s2B)4.7 m/s2C)6.2 m/s2D)5.2 m/s22)F = mgsin(angle); f = mu*m*g cos(angle); a = (F-f)/m = g (sin(angle)-mu*cos(angle))3)A skydiver reaches a "terminal velocity" of 120 km/h. If the skydiver has a mass of 89.0 kg,what is the magnitude of the upward force on the skydiver due to wind resistance?A)9.08 NB)8.17 NC)872 ND)959 N3)Upward force = mg since there is no acceleration.4)A driver in a 1000.0 kg car traveling at 37 m/s slams on the brakes and skids to a stop. If thecoefficient of friction between the tires and the road is 0.80, how long will the skid marks be?A)109 mB)70 mC)87 mD)81 m4)deceleration, a = mu*g; x = (v*v/2*a)5)A rescue plane spots a survivor 123 m directly below and releases an emergency kit with aparachute. If the package descends at a constant vertical acceleration of 7.09 m/s2and theinitial plane horizontal speed was 70.9 m/s, how far away from the survivor will it hit thewaves?A)2.46 kmB)436 mC)418 mD)296 m5)calculate t and then x using y = 1/2*a*t*t; x = v*t6)A robot submersible is released from a research vessel. Through computer controls the craft isto execute the following sequence: a) a = 3.18 i^ - 3.60 j^ m/s2 for 24 s, b) maintain its velocity(no acceleration) for another 4.05 min, and c) come to a full stop. How far from the vessel willit be located?A)1200 i^ - 1400 j^ mB)2100 i^ - 2400j^ mC)1200 i^ + 1400 j^ mD)1100 i^ - 1200 j^ m6)s = 1/2*a*t1*t1 + a*t1*t2; where a = acceleration vector, t1 = 24 s, t2 = 4.05 sec (not 4.05 min!)7)A boy throws a rock with an initial velocity of 2.30 m/s at 30.0° above the horizontal. How longdoes it take for the rock to reach the maximum height of its trajectory?A)0.117 sB)0.207 sC)0.324 sD)0.230 s7)t = v*sin(30deg)/g8)A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocitywas 7.22 m/s, what is the highest point of its trajectory?A)2.00 mB)1.30 mC)28.88 mD)4.00 m8)t = v*sin(60deg)/g; y = v*t + 1/2*g*t*t19)A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter. If themerry-go-round makes 8.4 rev/min, what is the velocity of the child in m/s?A)15.8 m/sB)7.9 m/sC)5.5 m/sD)1.3 m/s9)v = omega*r = omega*d/2, where d = diameter = 2*r10)A satellite is in orbit around a planet. The orbital radius is 29.0 km and the gravitationalacceleration at that height is 3.7 m/s2. What is the satellite's orbital speed?A)10 m/sB)330 m/sC)33 m/sD)100 m/s10)v = square-root of (a*r)11)A 22 kg mass is connected to a nail on a frictionless table by a (massless) string of length 1.3 m.If the tension in the string is 51 N while the mass moves in a uniform circle on the table, howlong does it take for the mass to make one complete revolution?A)4.7 sB)4.4 sC)3.8 sD)5.1 s11)Find omega from this: m*omega^2*r = tension, and use omega here: T= 2*pi/omega12)You are taking a turn at 39.0 m/s on a ramp of radius 29.0 m. What is your acceleration?A)0.744 m/s2B)52.4 m/s2C)1.34 m/s2D)21.6 m/s212)a = v^2/r13)An aircraft performs a maneuver called an aileron roll. During this maneuver, the plane turnslike a screw as it maintains a straight flight path, by using its ailerons to set the wings incircular motion. If it takes it 39 s to complete the circle and each wing length is 4.5 m, what isthe acceleration of the wing tip?A)0.54 m/s2B)8.3 m/s2C)1.9 m/s2D)0.12 m/s213)omega = 2*pi/T; a = omega^2*r14)In an amusement park ride passengers stand inside an 8 m radius cylinder. Initially thecylinder rotates with its axis oriented along the vertical. After the cylinder has acquiredsufficient speed, it tilts into a vertical plane, that is, the axis tilts into the horizontal, as shownin the figure. Suppose that, once the axis has tilted into the horizontal, the ring rotates onceevery 4.5 s. If a rider's mass is 44 kg, with how much force does the ring push on her at the topof the ride?A)260 NB)430 NC)690 ND)1100 N14)F = m*omega^2*r, where omega = 2*pi/T15)Two boxes are sitting side by side on a frictionless surface. The box on the left has a mass of11 kg, and the box on the right has a mass of 17 kg. If a 28 N force pushes on the 11 kg boxfrom the left, what is the force exerted on the 17 kg box by the 11 kg box?A)17 NB)21 NC)11 ND)14 N15)This force gives acceleration to both: F = (m1+m2)a; F-F21 = m1*a; solve for F212Solve the problem. (Use g = 9.8 m/s2.)16)A 10.0 kg block on a table is connected by a string to a 63 kg mass, which is hanging over theedge of the table. Assuming that frictional forces may be neglected, what is the magnitude ofacceleration of the 10.0 kg block when the other block is released?A)8.5 m/s2B)8.1 m/s2C)9.0 m/s2D)7.5 m/s216)a = m2*g/(m1+m2) from the equations: T = m1*a; m2*g-T = m2*a17)A 12 kg block on a table is connected by a string to a 26 kg mass, which is hanging over theedge of the table. If the 12 kg block is 2.0 m from the edge of the table, how much time willpass before the block falls off the table from when the other block is released? Assume thatfrictional forces may be neglected.A)0.65 sB)0.46 sC)0.77 sD)0.55 s17)Find a using the same idea as in question #16, and use the equation of kinematics: x = 1/2*a*t^218)A piano mover raises a 100 kg piano at a constant rate using a frictionless pulley system, asshown below. With roughly what force is the mover pulling down on the rope?A)1000 NB)2000 NC)500 ND)250 NE)Depends on the velocity!18)2*T = mg and T = mg/2319)The figure shows two packages that start sliding down a 20° ramp from rest a distanced = 3.1 m along the ramp from the bottom. Package A has a mass of 5.0 kg and a coefficient offriction 0.20. Package B has a mass of 10 kg and a coefficient of friction 0.15. How long does ittake package A to reach the bottom?A)2.0 sB)2.2 sC)2.4 sD)1.8 s19)Forces on A: Fa - fa + Fba = m1*a; Forces on B: Fb-fb-Fab = m2*a; Fba = -Fab (Newton's 3rdlaw)Fa = force down the plane for A =