Physics Diagnostic 1A
The earth's orbit around the sun is very nearly a perfect circle. The distance from the earth to the sun is 1.5 × 1011 meters, and one earth year is about π × 107 seconds. What is the magnitude of the earth's acceleration toward the sun? A. 3.5 × 1022 m/s2 B. 1.8 × 105 m/s2 C. 6 × 104 m/s2 D. 6 × 10-3 m/s2
D. Because the orbit is circular, the acceleration experienced by the earth is centripetal. Using r as the radius and t as the period yields: . 6E-3 m/s^2
Initial velocity and instantaneous acceleration are as shown: which of the following is true about the velocity one second later? A. The speed is greater and the velocity has turned slightly to the right. B. The speed is less and the velocity has turned slightly to the right. C. The speed is greater and the velocity is now in the direction of a. D. The speed is the same and the velocity is now in the direction of a.
A. Acceleration is the change in velocity over time. Therefore, a can simply be added to v0 to get final velocity. Since the acceleration has a positive component parallel to the initial velocity, the speed must increase. This eliminates choices B and D. Choice C can be eliminated without drawing a vector diagram: since v0 has some initial direction, the velocity at a later point cannot be in the direction of a (in general the sum of two vectors in different, non-opposite directions cannot be in the same direction as either of the vectors).
Jose asks Jill to slide the salad bowl (m = 2 kg) across the table. Jill's initial push gives it an initial speed of 4 m/s, and it comes to a halt directly in front of Jose after traveling 2 meters. What was the magnitude of the kinetic frictional force slowing the bowl during its slide? A. 8 N B. 4 N C. 2 N D. 0.8 N
A. First, consider what is known and what is desired. Eventually, force is desired. One of the ways to determine force is through Newton's second law, which requires mass and acceleration (the force of friction cannot be determined directly with its equation, as such values are not given). Since the mass is given, the acceleration is needed to determine the force. The question provides enough details to determine acceleration using kinematics: . Once a is determined, Fnet = ma can be used to determine force. Since the frictional force is the only force acting in the plane of motion Fnet = ma = 2(-4) = -8 N. Since all answer choices are positive and only magnitude is of interest, the negative sign can be ignored. Note: this question could also be solved using the work-energy theorem.
A car makes a left turn (radius of curvature of 20 m) on a flat road at a constant speed 10 m/s. If the car has a mass of 1000 kg, what minimum static friction coefficient is required between the tires and the road to maintain the turn? A. 0.25 B. 0.5 C. 1 D. Cannot be determined from the information given
B. In this case, static friction provides the centripetal acceleration (because the tires are not slipping on the road). Therefore: 0.5
A 10 kg, 4 m long plank of wood is going to be used as a teeter-totter for a brother and sister. The brother has a mass of 30 kg and the sister a mass of 20 kg. If the brother and sister sit at opposite ends of the plank, how far from the brother should the fulcrum be in order for the teeter-totter to be balanced? A. 1.33 m B. 1.60 m C. 1.67 m D. 2 m
C. To solve this problem, we use the equation for the center of mass in one dimension, remembering to include a term for the mass of the plank, M, at its geometric center (i.e. the midpoint of the plank). Setting the location of the brother as the origin (x1 = 0), we see that:
How much kinetic energy does a 1500 kg car have when it is moving at a speed of 20 m/s? A. 6 × 105 J B. 3 × 105 J C. 1.5 × 105 J D. 3 × 104 J
B. 3E5 J
It takes light from the sun 8 minutes to reach the earth. If light travels 3 × 108 m/s, how far is the sun from the earth in meters? A. 2.4 × 109 m B. 1.44 × 1010 m C. 1.44 × 1011 m D. 2.4 × 1011 m
C. Solving this requires multiplying the speed of light by eight minutes. Since the speed of light is in m/s, the seconds must be converted to minutes, as seen below: 1.44E11
An ideal projectile launched from the ground at an angle of 45° up from the horizontal has a total flight time of T. Which of the following correctly expresses a relationship between the displacement d, velocity v, and acceleration a at T / 2 compared to time T (the instant before it hits the ground)? A. 2 × |dT/2| > |dT| B. |vT/2| = |vT| C. |aT/2| > |aT| D. v T/2 = 0
A. At half the total flight time, a projectile launched upward from even ground upon which it will eventually land will be at the apex of its flight. At this time, all of its velocity is in the horizontal direction (meaning it has slowed down, eliminating choice B, but is still moving, eliminating choice D). Its acceleration throughout the flight is a constant g downward, eliminating choice C. That only leaves answer choice A, which is correct: The magnitude of the displacement from the launch point to the apex is a diagonal line that is the hypotenuse of a right triangle with the max height of the projectile as one leg and half the range (R = |dT|) as the other. Twice this hypotenuse must therefore be greater than the total range dT.
Consider the pulley system below. Find the acceleration of mass m2 up the hill.
A. Since the question states that m2 will accelerate up the hill, the direction of rotation of the pulley will be counterclockwise, meaning that down will be positive for m1 and up the inclined plane will be positive for m2. Here then are the two tension equations: . Note that the force down the plane has the sine component (the force into the plane and normal force have the cosine component). Adding the two equations yields: m1g - m2g sinθ = m2a + m1a, or (m1 - m2 sinθ)g = (m1 + m2)a. Solving for acceleration yields:
Tired from studying too much physics, Arthur pushes his face into the wall. If the force Arthur's stationary face exerts on the wall is Fpush and the force of the wall on Arthur's face is Fnormal, what is true about these two forces? A. They are equal and opposite and an action-reaction pair. B. They are equal and opposite but not an action-reaction pair. C. They are not equal in magnitude but are an action-reaction pair. D. They are not equal in magnitude and are not an action-reaction pair.
A. The pushing force of the face on the wall is equal and opposite to the normal force of the wall on the face. Since they are two equal and opposite forces of the same type between two objects (fulfilling Newton's third law), they are indeed an action-reaction pair.
A bobsled team of four men, each of mass m, pushes the sled of mass M over a horizontal displacement d before leaping into the sled and beginning to slide down the hill. The total length of the winding track from the top of the hill to the bottom is L, and the height difference between the top of the hill and the bottom is Δh. If the sled starts at rest and we assume the whole track is frictionless, how does the final kinetic energy of the sled at the bottom of the hill (prior to the brakes being applied) compare to the initial potential energy of the sled at the top of the hill? A. KE f > PEi B. KE f = PEi C. KE f < PEi D. It is not possible to uniquely determine the relationship between these two energies.
A. With no friction, this is a case of conservation of mechanical energy after work is done on the combined mass by the running men. The conservation relation can be written KEfinal = PEinitial + Wexternal or, if you take the speed achieved by the bobsled team just as they leap into the sled, KEfinal = PEinitial + KEinitial. Either way it is clear that the final kinetic energy exceeds the initial potential energy of the sled.
The mechanical advantage of a machine is a ratio of the force required to lift an object a given distance to the force required to move that object the same vertical distance using a simple machine. A ramp is one example of a simple machine that allows one to exert less force in moving an object from a lower to a higher point. Suppose a box is pushed along a 5 meter long ramp with a rise of 2 meters (with negligible friction between the box and the ramp). What mechanical advantage is gained by using the ramp instead of just lifting the box straight up? A. 5 B. 2.5 C. 0.2 D. no advantage
B. Because the work involved is the same whether the mass is lifted straight up or pushed up the ramp, mechanical advantage can also be calculated as the ratio of the distance over which the force is exerted using the simple machine to the vertical displacement: . Thus in this case MA = 5/2 = 2.5.
A person rides a bike uphill at a constant speed. What best describes the work and energy changes within the system? A. There is no work being done by the person because the speed isn't changing. B. The person is doing external work on the system that increases the gravitational potential energy. C. The nonconservative work being done by kinetic friction on the bike tires cancels out the positive external work the person does in pumping the pedals. D. It is impossible to say how much work is being done without knowing the angle of inclination of the hill.
B. External work done on a system in which mechanical energy is conserved increases either the kinetic or the potential energy (or both) of the system. There are no nonconservative forces like kinetic friction (since the tires aren't slipping), so mechanical energy is conserved; with a constant speed, kinetic energy isn't changing, so the increase in potential energy for the uphill movement must be accomplished by the work done on the system by the person peddling the bike.
A book (m = 1 kg) falls off of the top shelf (h = 1.5 m) of a bookcase and hits the floor. It is then accidentally kicked across the floor, moving a distance of 3 m before friction (µk = 0.25) brings it to a stop. The book is then returned to the top shelf 2 m to the right of its original position. What is the change in the gravitational potential energy of the book during this process? A. 20 J B. 0 J C. -15 J D. -22.5 J
B. Gravity is a conservative force, so the change in gravitational potential energy depends only upon the initial and final heights of the book, not on the path the book takes to get from one to the other: ΔPEgrav = mgΔh. Because there is no change in height in moving from the top shelf back to the top shelf, there is no change in energy.
A person walks east along a road. As he steps forward, the foot in contact with the ground experiences which of these forces? A. A static frictional force west B. A static frictional force east C. A kinetic frictional force west D. A kinetic frictional force east
B. Process of elimination is useful here. Choices C and D are wrong because unless the individual's feet are slipping on the road, there is no kinetic friction (remember not to oversimplify: kinetic friction doesn't apply to all moving objects, only slipping objects). The question then is this: In which direction does the static frictional force on the person's feet act? As the individual walks to the east, his/her feet when in contact with the road push back to the west. In other words, if his/her feet were to slip, they would slip backwards, the opposite of the direction the individual is walking. Put another way, if a man goes running on a dirt track, the dust and sand he kicks up shoots out behind him, not in front of him, because toward the back is the direction of force the man's feet exert on the road. By Newton's third law, this implies that the static frictional force the road exerts on the man's feet is equal and opposite, meaning in the direction of the man's motion.
A ball is tossed straight up in the air off of a bridge with an initial speed of 5 m/s. If it hits the water below the bridge with a speed of 15 m/s, from what height above the water was it tossed? A. 5 m B. 10 m C. 15 m D. 30 m
B. The method for free fall problems is the same as for other one-dimensional kinematics problems, with the added piece of information that acceleration is constant in the downward direction at about -10 m/s2. Note that the problem implicitly defines up as the positive direction by giving a positive initial speed upward. Thus, consider what is known and missing, and which equation matches that information: y=-10 The negative value of y indicates that the displacement of the ball is in the negative direction, since the upwards initial velocity was defined as positive (remember to be consistent with a positive direction). The height is the absolute value of this displacement.
A man drags a wooden crate (m = 50 kg) across the ground (µk = 0.2) with a rope at an angle of 45° up from the horizontal. If the tension in the rope is 200 N, how much work is done by the pulling force in moving the crate over a displacement of 5 m? A. 1000 J B. 700 J C. 340 J D. 200 J
B. The work done by a single force is product of the parallel components of that force and the displacement vector: W = Fd cosθ. Therefore WT = (200)(5)cos 45° ≈ (200)(5)(0.7) = 700 J.
A car moving 20 m/s in the positive x-direction slams on the brakes and stops after skidding 20 m. What was its acceleration during this period? A. -20 m/s2 B. -10 m/s2 C. -5 m/s2 D. -2 m/s2
B. This is a uniformly accelerated motion problem, which can be solved using the standard set of 5 kinematics equations. The technique is always the same: determine the applicable equation using the given and desired variables. In this case, the following variables are known or sought:. Those match Big Five #5, which is also indicated by the lack of time t: a=-10 m/s^2
Which of the following is a description of a system in translational equilibrium? A. A ball rolling around a circular track at constant speed. B. A car driving at a constant speed up a straight hill with an angle of inclination θ. C. An object in free fall. D. Two unequal, free-hanging masses attached by a rope running over a vertical pulley (i.e. an Atwood machine).
B. Translational equilibrium means the net force equals zero, so acceleration equals zero, which means that the velocity must be constant. A ball rolling at a constant speed but in a curved path does not have constant velocity, because direction is changing (eliminating choice A). An object in free fall is accelerating at g (eliminating choice C), and an Atwood machine in free motion has an acceleration of (m2 - m1)g / (m1 + m2) (eliminating choice D). Note that calculating the actual acceleration value is unnecessary since the acceleration must be nonzero so long as the masses are unequal and free to move.
The equation for the Reynolds number, which determines whether the flow of a fluid is laminar or turbulent, is R = ρvL / µ, where ρ is density, v is the flow speed, L is length, and µ is the dynamic viscosity of the fluid (measured in kg / m·s). What are the units of the Reynolds number? A. 1 / m B. unitless C. kg / m D. 1 / m·s
B. Unit analysis yields: 1
The moment of inertia I is the rotational analog of mass found in the rotational form of Newton's second law of motion, τ = Iα (where τ is the net torque on an object and α is its angular acceleration). The value of I depends upon the mass of an object and the distribution of that mass around the axis of rotation, so that the further the mass is from the rotation axis, the greater the value of I. Suppose you have a quarter (a flat disk) balanced on edge, a plain wedding band (a uniform ring) also on edge, and a marble (a solid sphere) that you are going to set spinning on a frictionless surface. Each object has the same mass and the same radius. If you apply the same torque to each object, which object will experience the least angular acceleration? A. The marble B. The quarter C. The wedding band D. They will all experience the same angular acceleration.
C. A ring is hollow and therefore has a greater proportion of its mass concentrated further from the axis of rotation than either of the other objects, eliminating choices A and B (and by extension, choice D). The difference between the quarter and the marble is harder to perceive without doing the math to compute the moments of inertia, but you would be given these formulae on the MCAT if you needed them.
A man is using a rope tied to the top of a wooden crate of mangos to drag the crate up a long straight hill (angle of inclination θ, coefficient of kinetic friction µk > 0). As the man tires, the speed of the crate slows. As the crate is slowing, what is true of the vector sum of the normal force, weight, tension, and kinetic friction forces? A. The net force points up the hill. B. The net force is zero. C. The net force points down the hill. D. The net force points straight down.
C. According to Newton's second law, the direction of the net force corresponds to the direction of the acceleration. Because the crate is moving up the hill but slowing down, the acceleration points down the hill. Thus the sum of the forces acting on the crate must do the same.
A book of mass 1 kg is pushed across a desk in the +y direction and released. After one second, a kinetic frictional force of 10 N causes the book to slide to a stop. During that one second, what net force did the book exert on the table? A. A normal force straight down of 10 N B. A diagonal force up and in the -y direction of 14.1 N C. A diagonal force down and in the +y direction of 14.1 N D. A diagonal force down and in the -y direction of 14.1 N
C. As a book moves across the table after being released, three forces act on it. Weight pulls it down into the table, the normal force pushes it up from the table, and kinetic friction pushes it opposite its direction of motion (so in the -y direction). Only the latter two of these forces are contact forces with the table, so only those two have equal and opposite reaction forces that act from the book on the table (note: Newton's third law dictates that both the individual forces and the sum of the forces between the book and the table are equal and opposite). The reaction force FN book on table is down, and fk book on table is in the +y direction. Their sum is diagonal (a composition of down and forward) and has a magnitude of . Note also that the math isn't necessary to discern between the answer choices: discerning the direction will eliminate all but choice C.
Due to frictional effects within the car and air drag, a car's tires must exert a force of 104 N on the road to maintain a constant speed of 67 mph (30 m/s). What power is being transferred by the transmission to the tires at this speed? A. 0 W B. 3 × 104 W C. 3 × 105 W D. 6.7 × 105 W
C. For a constant force parallel to velocity, power is given by P = Fv. In this case that yields P = 104 N × 30 m/s = 3 × 105 W.
Torque has which of the following properties? It is measured in joules. It is a vector. It can be zero when force is nonzero. A. I only B. I and II C. II and III D. I, II, and III
C. I is false: torque is measured in Newton-meters (N•m). While dimensionally equivalent to joules, they are not considered the same since joules measure scalar quantities such as work and energy. This eliminates choices A, B and D. II is true: torque is indeed a vector, though its vector properties are not typically tested on the MCAT. This eliminates choice A. III is also true: an object can have a force acting on it without exerting a torque if that force's line of action passes through the center of mass of the object (you can think of this as the case in which the angle between the radial vector from the pivot point to where the force is applied and the force vector is 0° or 180°, so sine = 0). This eliminates choices A and B.
A block with a mass of 5 kg starts at rest at the higher end of a 5 meter long tilted ramp. The difference in height between the high and low ends of the ramp is 3 meters. The ramp itself is smooth enough that friction can be ignored as the block slides down, but once the block reaches the ground, it experiences a kinetic frictional force as it slides to a stop. If the block slides to a stop in 6 meters on the horizontal ground, what is µk between the block and the ground? A. 0.83 B. 0.67 C. 0.5 D. 0.33
C. Kinetic friction does the nonconservative work to bring the sliding block to a stop after its initial gravitational potential energy has been transformed into kinetic energy. Starting from the total energy conservation equation, the value of the coefficient of kinetic friction can be found from the work expression. Note that friction works over the horizontal ground, so the normal force is just mg, and that the block begins and ends at rest, so ΔKE = 0: Mk=0.5
A block with a mass of 5 kg rests on a horizontal table (µs = 0.3, µk = 0.2). A constant horizontal pushing force of 20 N is applied. How far does the block move in 2 seconds? A. 0 m B. 2 m C. 4 m D. 20 m
C. Like many friction problems, this is an example of a dynamics leading to kinematics problem. Begin by confirming that the block does in fact move, which can be done by determining the maximum static frictional force: Fs max = µsmg = (0.3)(5)(10) = 15 N. That's less than the applied force, indicating that the block does indeed accelerate across the table, with kinetic friction working against this sliding motion. Next, the net force is needed, which can be found using Newton's second law. Fx net = Fapplied - Fkinetic friction = Fapplied - µkmg = 20 - (0.2)(5)(10) = 20 - 10 = 10 N. The acceleration can be determined. a = Fx net / m = 10 / 5 = 2 m/s2. Finally, kinematics can be used to find the distance. a = 2 m/s2; v0 = 0 m/s; t = 2 s; d = ? so v is missing and d = v0t + 0.5at2 = 0 + (0.5)(2)(2)2 = 4 m.
Two blocks start at height h: block 1 on a frictionless ramp of constant angle of inclination, block 2 suspended in midair. Both blocks are released at the same moment. How does the average power exerted on the blocks during their descent compare? A. P 1 > P2 B. P 1 = P2 C. P 1 < P2 D. Cannot be determined based on the information given.
C. Power is work / time. The work done by gravity on each block is the same, because the change in height of each block is the same. The normal force on block 1 does no work because it is perpendicular to displacement. Since there is no friction on the plane, the net work done on each block must therefore be the same. The time it takes block 2 to descend over the displacement h is less than the time it takes block 1 to displace a longer distance down the ramp (h / sinθ), both because block 2 undergoes less displacement and because block 2 experiences greater acceleration (g versus g sinθ). Less time for the same work means more power.
Which of the following is true about the normal force? A. It is always equal and opposite to the weight. B. It can cause an object moving along a surface without colliding to change speed. C. It is attributable to the binding forces between atoms and molecules in solid matter. D. It cannot contribute to the change in direction of a moving object.
C. Process of elimination is the best way to approach this question, unless you happen to recognize choice C as being true at first reading. Choice A is false because the normal force is always perpendicular to the plane of contact between two bodies, whether that plane is horizontal or not (a box on a hill experiences a normal force that is less than and not in the opposite direction to the weight). Because normal forces are perpendicular to the direction of motion of an object moving over the contact surface, they do no work on such an object and therefore do not change the kinetic energy or speed (eliminating choice B). A counterexample to choice D is a banked circular track like a racing oval, where the normal force provides some of the centripetal acceleration of the car going around the track. The reason one does not fall through the floor is indeed because the various electromagnetic bonds resist the breaking that would occur if one's foot passed through the ground.
As an object moves horizontally across a surface, it is subject to two horizontal forces. The magnitude and direction of those two forces is graphed as a function of the displacement of the object, as shown below (one force is indicated by the solid line, the other by the dashed line). What is the net work done on the object over its displacement? A. 5 J B. 0 J C. -5 J D. -10 J
C. The area under the force versus displacement curve gives the work done by that force, and the net work is the sum of the works done by all the relevant forces. The area under the positive curve is the area of a triangle with a base of 2 m and a height of 5 N: one half base times height gives +5 J. The area under (or "over") the negative force curve is the area of a rectangle with a base of 5 m and a height of -2 N, or -10 J. 5 - 10 = -5 J.
For which of the following objects does the center of mass fall at a point where none of the object's matter is located? A. A metal disk B. A solid plastic cube C. A golden ring D. A sheet of paper
C. The center of mass of an extended body (a mass in two or three dimensions) of uniform density is the geometric center of that body. For a cylinder, this means the center of the circular cross section at the midpoint of the height of the cylinder. Because a ring is a hollow cylinder, there is no ring material at this point. Each of the other examples has material at its geometric center.
An object accelerates uniformly from rest and moves a distance X meters in time T seconds. If the object accelerates uniformly from rest with the same acceleration, how far does it move in time 2T seconds? A. (1.4)X meters B. 2X meters C. 4X meters D. Impossible to determine with the given information.
C. The first sentence gives the variables d, v0, and t, and from this we could find either a or v. One option would be to solve for a and use that with the information given in the second sentence to find d for t = 2T. However, it is more efficient simply to note that with v0 = 0, d ∝ t2, since d = v0t + 0.5at2. Hence, if time doubles (with all other conditions being equal), distance must quadruple.
A mass m is released from rest at the top of a frictionless inclined plane with a 3 meter base and a 4 meter height. With what speed does the mass reach the ground at the base of the plane? A. 4 m/s B. 5 m/s C. 9 m/s D. 11 m/s
C. The force down an inclined plane due to gravity is Fdown plane = mg sinθ, so a = g sinθ, v0 = 0, and d = 5 m (because the ramp is the hypotenuse of a 3-4-5 right triangle). Thus, kinematics can be used to find the final speed (in this case time is missing): v=9m/s . Note, this can also be solved using conservation of energy.
Three forces act on an object as shown in the free-body diagram. The forces F1 and F2 are perpendicular to one another, and F3 is straight down. Which of the following correctly expresses the net horizontal force acting on the object? A. F 1 - F2 B. F 1 cosθ - F2 cosθ C. F 1 sinθ - F2 cosθ D. F 1 sinθ - F2 sinθ
C. The net force in the horizontal direction is composed only of the horizontal components of F1 and F2. The horizontal component of F1 is opposite the angle θ, making it F1 sinθ. Because the angle between F1 and F2 is 90°, the angle between F2 and the vertical is 90° - θ, which means the angle between F2 and the horizontal is θ (since the vertical and horizontal lines are perpendicular). Thus the horizontal component of F2 is F2 cosθ. Finally, the net force in the horizontal direction is the sum of those two horizontal components. Plugging in the forces (remember the direction) yields Fnet = F1 sinθ - F2 cosθ.
Two pulleys--one mounted in the ceiling, another anchored to a mass M sitting on a platform below--have a rope looped over them three complete times, so that there are six strands of rope running between the two pulleys. One end of the rope is tied to the ceiling adjacent to the top pulley, and the other is being held by a man standing next to the mass. The man pulls down with a tension T on that strand of rope. If the mass just barely lifts off the ground, what is true of T? A. T = Mg B. T = 6Mg C. T = Mg / 6 D. T = Mg / 3
C. The pulleys create a mechanical advantage by multiplying the tension force by the number of strands running up from the bottom pulley. A pulley is simply a means of changing the direction of the tension force: it does not change its magnitude, nor does it change the basic property of ropes or strings, namely that they cannot push. This means that every strand connected to the bottom pulley applies a pulling force upward of magnitude T, so 6T = Mg (the weight of the mass being the minimum applied force upward required to lift it). Thus T = Mg / 6.
A bicycle tire has a radius of 33 cm. If the bicycle is traveling at 12 m/s, with what approximate frequency are the tires rotating? A. 0.6 rotations/sec B. 3 rotations/sec C. 6 rotations/sec D. 12 rotations/sec
C. The relationship between translational velocity and rotational frequency depends upon how far the bike moves for each complete rotation of the tires, i.e., on the circumference of the tire. Frequency is the reciprocal of the period, which is the time it takes to complete one revolution. Since period = 2πr / v, then frequency is equal to 6 rotations/sec
The slope at a point on a velocity versus time graph represents what physical quantity? A. Displacement B. Instantaneous velocity C. Instantaneous acceleration D. Average acceleration
C. The slope of a curve is rise over run. In this case, rise (y) is velocity and run (x) is time, yielding velocity / time or acceleration. This eliminates choices A and B. If the slope is measured at a particular point instead of between two points, the quantity represented is instantaneous, not average, eliminating choice D.
A soccer ball is kicked at a speed of 30 m/s at an angle of 30° up from the horizontal. How far will it travel horizontally before landing again on the field? A. 45 m B. 60 m C. 79 m D. 150 m
C. To answer horizontal range questions, start by determining the time of flight. For that, focus on the vertical component of motion, picking a location where there is enough information to solve for the time. This occurs at the apex: . Given the time of flight, the range is a matter of distance = rate × time, where the rate is the horizontal, cosine component of velocity and total time is twice tapex: R = 30 m/s (cos30°) × 3 s ≈ 90 m (.85) ≈ 77 m, closest to choice C.
A wooden plank in the shape of a rectangle with side lengths a and b lies on a frictionless surface. Two forces, each of magnitude F, are applied to opposite corners as shown. What is the magnitude and direction of the net torque applied to the plank about its center? A. aF / 2 counterclockwise B. bF / 2 counterclockwise C. aF clockwise D. bF clockwise
D. From the picture, it is clear that each force provides a clockwise force about its center. This eliminates choices A and B. For two dimensional bodies like the rectangular plank, the preferred formula for torque is the moment arm equation, τ = lF, where l, the moment arm (or lever arm), is the component of r that is perpendicular to the force. Another way of thinking about the moment arm is that it is the distance from the pivot to the line containing the force. In this case, l = b / 2 for both forces. Since each torque is in the same direction, the net torque is twice the torque provided by either force.
A man pushes down at an angle of 60° from the horizontal on an initially stationary 100 kg crate on a hockey rink floor (no friction) with a force of 200 N. If the man applies this force for 3 seconds, how far does the crate move? A. 450 m B. 9 m C. 6 m D. 4.5 m
D. One method of solving this question is to use Newton's second law to find the acceleration of the crate and then apply kinematics equations to solve for distance (this move from the dynamics of F = ma to kinematics is common). The crate must accelerate in the horizontal direction given the set up of the problem. Starting with Newton's second law and plugging in the known values yields: . Using this ax as a kinematic variable now yields: 4.5 m
Which of the following kinematic descriptions is physically impossible? A. An object in motion for some period of time has zero average velocity but positive average speed. B. An object maintains constant speed but has nonzero acceleration. C. An object in motion has zero displacement but positive total distance traveled. D. An object in motion for some period of time has an average velocity with a greater magnitude than its average speed.
D. This question can be answered using process of elimination. Choices A and C are both possible (and therefore eliminated as answer choices) because an object traveling in a closed curved path (like a circle) will have zero displacement and therefore zero average velocity but not zero distance or average speed. Choice B is possible, therefore eliminated, because it is the case for uniform circular motion. Hence, choice D must be correct. Average velocity has magnitude of displacement / time, whereas average speed equals total distance / time. The time is the same for both, and displacement ≤ total distance, so average velocity ≤ average speed. Put another way, there's no shorter path than a straight line, for which the magnitude of displacement equals the total distance. For any other type of path, the displacement is shorter than the distance traveled.
A shark is swimming north at a constant speed. What must be true about the propulsive force of the shark's swimming motions and the drag force of the water pushing back on the shark if these are the only two forces with components acting parallel to the direction of motion? A. The two forces are equal in magnitude. B. The magnitude of the propulsive force is greater than the drag force because propulsion has to overcome drag for the shark to move forward. C. The magnitude of the drag force is greater than the propulsive force because it is keeping the shark from accelerating. D. It is not possible to specify the relationship between these two forces without knowing the relations among the other forces acting on the shark.
A. Newton's first law states that any object moving at a constant velocity must have zero net force acting on it. Since the only two forces acting in the direction of motion are the propulsive and drag forces, these forces must be equal and opposite if they sum to zero.
A lever is used to lift a 10 kg mass to a height of 2 m. If the person on the other side of the lever exerted a force of 50 N over a vertical distance of 5 m to accomplish this lift, what is the efficiency of the lever system? A. 125% B. 80% C. 12.5% D. 8%
B. Efficiency is the percentage of work output by a machine relative to the work input or energy consumed. In this case, the lever lifted a 10 kg mass to a height of 2 m, for a work output of mgh = 200 J. Since the person input W = Fd = (50 N)(5 m) = 250 J, the efficiency is 200 / 250 = 0.8 = 80%.
A car of mass m drives around a circular banked track of radius R at constant speed v, and maintains a constant height h. What best describes the work done by the normal force on the car during one quarter of a revolution around the track? A. WN < 0 B. WN = 0 C. WN > 0 D. The work done by the normal force can be positive or negative depending upon the direction the car is moving.
B. Work done by a force on an object depends upon the component of the force along the direction of the displacement of the object. Anything moving in a circular path is, at any given instant, displacing in a direction tangent to the path (the instantaneous velocity), while it is accelerating along the radius toward the center of the circle (centripetal acceleration). Thus the force and displacement are perpendicular, and so the work done by the force is zero. It is also useful to consider that, because the car is not changing height or speed, it is neither gaining nor losing mechanical energy, so no work is being done on it by an outside force.
A book sits at rest on a horizontal table. It is accelerated with a pushing force. Then, after it has moved a distance s, it is released. Soon after it comes to rest. From the beginning to the end of this problem, what is the net work done on the book? A. Fpush × s B. -Fpush × s C. -Ffriction × s D. 0 J
D. The net work done on the book as it moves across a horizontal surface (over which there is no change in gravitational potential energy) must equal the change in kinetic energy. Because the book started and ended at rest, ΔKE = 0 = Wnet.