1. Question
A ball of mass m is suspended from two strings of unequal length as shown above. The magnitudes of the tensions T1 and T2 in the strings must satisfy which of the following relations?
(A) Tl = T2 (B) T1 > T2 (C) T1 < T2 (D) Tl + T2 = mg
Answer/Explanation
Ans:
C
Solution: As T2 is more vertical, it is supporting more of the weight of the ball. The horizontal components of T1 and T2 are equal.
Questions 2 –3
A 2-kg block slides down a 30° incline as shown above with an acceleration of 2 m/s2.
2. Question
Which of the following diagrams best represents the gravitational force W. the frictional force f, and the normal force N that act on the block?
Answer/Explanation
Ans:
D
Solution: Normal force is perpendicular to the incline, friction acts up, parallel to the incline (opposite the motion of the block), gravity acts straight down.
3. Question
Which of the following correctly indicates the magnitudes of the forces acting up and down the incline?
(A) 20 N down the plane, 16 N up the plane
(B) 4 N down the plane, 4 N up the plane
(C) 0 N down the plane, 4 N up the plane
(D) 10 N down the plane, 6 N up the plane
Answer/Explanation
Ans:
D
Solution: The component of the weight down the plane is 20 N sin θ. The net force is 4 N, so the friction force up the plane must be 4 N less than 20 N.
4. Question
When the frictionless system shown above is accelerated by an applied force of magnitude the tension in the string between the blocks is (A) F (B) 2/3 F (C) ½ F (D) 1/3 F
Answer/Explanation
Ans:
D
Solution: The force between objects is the applied force times the ratio of the mass behind the rope to the total mass being pulled. This can be derived from a = F/mtotal and FT = mbehind the ropea
5. Question
A ball falls straight down through the air under the influence of gravity. There is a retarding force F on the ball with magnitude given by F = bv, where v is the speed of the ball and b is a positive constant. The ball reaches a terminal velocity after a time t. The magnitude of the acceleration at time t/2 is
(A) Increasing
(B) Decreasing
(C) 10 m/s/s
(D) Zero
Answer/Explanation
Ans:
B
Solution: Since the ball’s speed is increasing from rest, the retarding force F is also increasing. The net force, which is the weight of the ball minus F, is thus decreasing. So the acceleration also must decrease. Time t /2 is before the constant-speed motion begins, so the acceleration has not yet decreased to zero.
6. Question
A block of weight W is pulled along a horizontal surface at constant speed v by a force F, which acts at an angle of θ with the horizontal, as shown above. The normal force exerted on the block by the surface has magnitude
(A) greater than W
(B) greater than zero but less than W
(C) equal to W
(D) zero
Answer/Explanation
Ans:
B
Solution: For vertical equilibrium, the weight equals the normal force plus the vertical component of F. This leads to the normal force being W – something. The block remains in contact with the surface, so the normal force does not reach zero.
7. Question
A uniform rope of weight 50 N hangs from a hook as shown above. A box of weight 100 N hangs from the rope. What is the tension in the rope?
(A) 75 N throughout the rope (B) 100 N throughout the rope
(C) 150 N throughout the rope (D) It varies from 100 N at the bottom of the rope to 150 N at the top.
Answer/Explanation
Ans:
D
Solution: The bottom of the rope supports the box, while the top of the rope must support the rope itself and the box.
8. Question
When an object of weight W is suspended from the center of a massless string as shown above, the tension at any point in the string is
(A) 2Wcosθ (B) ½Wcosθ (C) W/(2cosθ) (D) W/(cosθ)
Answer/Explanation
Ans:
D
Solution: The vertical components of the tension in the rope are two equal upward components of Tcosθ, which support the weight. \(\sum F_{y} = 0 = 2Tcos\theta – W\)
9. Question
A block of mass 3m can move without friction on a horizontal table. This block is attached to another block of mass m by a cord that passes over a frictionless pulley, as shown above. If the masses of the cord and the pulley are negligible, what is the magnitude of the acceleration of the descending block?
(A) g/4 (B) g/3 (C) 2g/3 (D) g
Answer/Explanation
Ans:
A
Solution: \(\sum F_{external} = m_{total}a;\) mg is the only force acting from outside the system of masses so we have mg = (4m)a
A plane 5 meters in length is inclined at an angle of 37°, as shown above. A block of weight 20 N is placed at the top of the plane and allowed to slide down.
10. Question
The magnitude of the normal force exerted on the block by the plane is
(A) greater than 20 N
(B) greater than zero but less than 20 N
(C) equal to 20 N
(D) zero
Answer/Explanation
Ans:
B
Solution: The weight component perpendicular to the plane is 20 N sin 37o. To get equilibrium perpendicular to the plane, the normal force must equal this weight component, which must be less than 20 N.
11. Question
Multiple correct: Three forces act on an object. If the object is moving to the right in translational equilibrium, which of the following must be true? Select two answers.
(A) The vector sum of the three forces must equal zero.
(B) All three forces must be parallel.
(C) The magnitudes of the three forces must be equal.
(D) The object must be moving at a constant speed.
Answer/Explanation
Ans:
A, D
Solution: (A) is the definition of translational equilibrium. Equilibrium means no net force and no acceleration, so (D) is also correct.
12. Question
For which of the following motions of an object must the acceleration always be zero?
(A) Any motion in a straight line
(B) Simple harmonic motion
(C) Any motion at constant speed
(D) Any single object in motion with constant momentum
Answer/Explanation
Ans:
D
Solution: Motion at constant speed includes, for example, motion in a circle, in which the direction of the velocity changes and thus acceleration exists. Constant momentum for a single object means , that the velocity doesn’t change.
13. Question
A rope of negligible mass supports a block that weighs 30 N, as shown above. The breaking strength of the rope is 50 N. The largest acceleration that can be given to the block by pulling up on it with the rope without breaking the rope is most nearly
(A) 6.7 m/s2 (B) 10 m/s2 (C) 16.7 m/s2 (D) 26.7 m/s2
Answer/Explanation
Ans:
A
Solution: \(\sum \) F = ma; FT – mg = ma; Let FT = 50 N (the maximum possible tension) and m = W/g = 3 kg
Questions 14-15
A horizontal, uniform board of weight 125 N and length 4 m is supported by vertical chains at each end. A person weighing 500 N is hanging from the board. The tension in the right chain is 250 N.
14. Question
What is the tension in the left chain?
(A) 125 N (B) 250 N (C) 375 N (D) 625 N
Answer/Explanation
Ans:
C
Solution: The sum of the tensions in the chains (250 N + Tleft) must support the weight of the board and the person (125 N + 500 N)
15. Question
Which of the following describes where the person is hanging?
(A) between the chains, but closer to the left-hand chain
(B) between the chains, but closer to the right-hand chain
(C) Right in the middle of the board
(D) directly below one of the chains
Answer/Explanation
Ans:
A
Solution: The board itself provides the same torque about the attachment point of both chains, but since the left chain provides a bigger force on the board, the person must be closer to the left chain in order to provide an equivalent torque on both chains by τ = Fd.
16. Question
Multiple correct: The cart of mass 10 kg shown above moves without frictional loss on a level table. A 10 N force pulls on the cart horizontally to the right. At the same time, a 30 N force at an angle of 60° above the horizontal pulls on the cart to the left. Which of the following describes a manner in which this cart could be moving? Select two answers.
(A) moving left and speeding up
(B) moving left and slowing down
(C) moving right and speeding up
(D) moving right and slowing down
Answer/Explanation
Ans:
A, D
Solution: The horizontal component of the 30 N force is 15 N left. So the net force is 5 N left. So the acceleration is left. This could mean either A or D – when acceleration is opposite velocity, an object slows down.
17. Question
Two people are pulling on the ends of a rope. Each person pulls with a force of 100 N. The tension in the rope is:
(A) 0 N (B) 50 N (C) 100 N (D) 200 N
Answer/Explanation
Ans:
C
Solution: Consider that no part of the system is in motion, this means at each end of the rope, a person pulling with 100 N of force is reacted to with a tension in the rope of 100 N.
18. Question
The parabola above is a graph of speed v as a function of time t for an object. Which of the following graphs best represents the magnitude F of the net force exerted on the object as a function of time t?
Answer/Explanation
Ans:
A
Solution: As v is proportional to t2 and a is proportional to Δv/t, this means a should be proportional to t
19. Question
A ball initially moves horizontally with velocity vi, as shown above. It is then struck by a stick. After leaving the stick, the ball moves vertically with a velocity vf, which is smaller in magnitude than vi. Which of the following vectors best represents the direction of the average force that the stick exerts on the ball?
Answer/Explanation
Ans:
B
Solution: The direction of the force is the same as the direction of the acceleration, which is proportional to Δv = vf + (–vi)
20. Question
Two massless springs, of spring constants k1 and k2, are hung from a horizontal support. A block of weight 12 N is suspended from the pair of springs, as shown above. When the block is in equilibrium, each spring is stretched an additional 24 cm. Thus, the equivalent spring constant of the two-spring system is 12 N / 24 cm = 0.5 N/cm. Which of the following statements is correct about k1 and k2?
(A) k1 = k2 = 0.25 N/cm
(B) 1/k1 + 1/k2 = 1/(0.5 N/cm)
(C) k1 – k2 = 0.5 N/cm
(D) k1 + k2 = 0.5 N/cm
Answer/Explanation
Ans:
D
Solution: A force diagram will show that the forces provided by each spring add up to 12 N: F1 + F2 = 12 N. Each force is kx; each spring is stretched the same amount x = 24 cm. So k1x + k2x = 12 N; dividing both sides by x shows that k1 + k2 = 0.5 N/cm.
21. Question
A ball is thrown and follows a parabolic path, as shown above. Air friction is negligible. Point Q is the highest point on the path. Which of the following best indicates the direction of the net force on the ball at point P ?
Answer/Explanation
Ans:
D
Solution: Net force is the gravitational force which acts downward
22. Question
A block of mass m is accelerated across a rough surface by a force of magnitude F that is exerted at an angle Φ
with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f. What is the acceleration of the block?
(A) F/m (B) (FcosΦ)/m (C) (F–f)/m (D) (FcosΦ–f)/m
Answer/Explanation
Ans:
D
Solution: \(\sum \) F = ma = FcosΦ – f
23. Question
Three blocks of masses 3m, 2m, ands are connected to strings A, B, and C as shown above. The blocks are pulled along a rough surface by a force of magnitude F exerted by string C. The coefficient of friction between each block and the surface is the same. Which string must be the strongest in order not to break?
(A) A (B) B (C) C (D) They must all be the same strength.
Answer/Explanation
Ans:
C
Solution: The string pulling all three masses (total 6m) must have the largest tension. String A is only pulling the block of mass 3m and string B is pulling a total mass of 5m.
24. Question
A block of mass 3 kg, initially at rest, is pulled along a frictionless, horizontal surface with a force shown as a function of time t by the graph above. The acceleration of the block at t = 2 s is
(A) 4/3 m/s2 (B) 2 m/s2 (C) 8 m/s2 (D) 12 m/s2
Answer/Explanation
Ans:
A
Solution: At t = 2 s the force is 4 N. F = ma
25. Question
A student pulls a wooden box along a rough horizontal floor at constant speed by means of a force P as shown to the right. Which of the following must be true?
(A) P > f and N < W.
(B) P > f and N = W.
(C) P = f and N > W.
(D) P = f and N = W.
Answer/Explanation
Ans:
A
Solution: Since P is at an upward angle, the normal force is decreased as P supports some of the weight. Since a component of P balances the frictional force, P itself must be larger than f.
26. Question
The 10.0 kg box shown in the figure to the right is sliding to the right along the floor. A horizontal force of 10.0 N is being applied to the right. The coefficient of kinetic friction between the box and the floor is 0.20. The box is moving with:
(A) acceleration to the left. (B) acceleration to the right.
(C) constant speed and constant velocity. (D) constant speed but not constant velocity.
Answer/Explanation
Ans:
A
Solution: The force of friction = μFN = 0.2 × 10 kg × 9.8 m/s2 = 19.6 N, which is greater than the applied force, which means the object is accelerating to the left, or slowing down
27. Question
Assume the objects in the following diagrams have equal mass and the strings holding them in place are identical. In which case would the string be most likely to break?
Answer/Explanation
Ans:
B
Solution: The upward component of the tension is Tup = Tsinθ, where θ is the angle to the horizontal. This gives T = Tup/sinθ. Since the upward components are all equal to one half the weight, the rope at the smallest angle (and the smallest value of sinθ) will have the greatest tension, and most likely break
28. Question
Two blocks of mass 1.0 kg and 3.0 kg are connected by a string which has a tension of 2.0 N. A force F acts in the direction shown to the right. Assuming friction is negligible, what is the value of F?
(A) 2.0 N (B) 4.0 N (C) 6.0 N (D) 8.0 N
Answer/Explanation
Ans:
D
Solution: From the 1 kg block: F = ma giving a = 2 m/s2. For the system: F = (4 kg)(2 m/s2)
29. Question
A 50-kg student stands on a scale in an elevator. At the instant the elevator has a downward acceleration of 1.0 m/s2 and an upward velocity of 3.0 m/s, the scale reads approximately
(A) 350 N (B) 450 N (C) 500 N (D) 550 N
Answer/Explanation
Ans:
B
Solution: Elevator physics: FN represents the scale reading. \(\sum \) F = ma; FN – mg = ma, or FN = m(g + a). The velocity of the elevator is irrelevant.
30. Question
A tractor-trailer truck is traveling down the road. The mass of the trailer is 4 times the mass of the tractor. If the tractor accelerates forward, the force that the trailer applies on the tractor is
(A) 4 times greater than the force of the tractor on the trailer.
(B) 2 times greater than the force of the tractor on the trailer.
(C) equal to the force of the tractor on the trailer.
(D) ¼ the force of the tractor on the trailer.
Answer/Explanation
Ans:
C
Solution: Newton’s third law
31. Question
A wooden box is first pulled across a horizontal steel plate as shown in the diagram A. The box is then pulled across the same steel plate while the plate is inclined as shown in diagram B. How does the force required to overcome friction in the inclined case (B) compare to the horizontal case (A)?
(A) the frictional force is the same in both cases
(B) the inclined case has a greater frictional force
(C) the inclined case has less frictional force
(D) the frictional force increases with angle until the angle is 90º, then drops to zero
Answer/Explanation
Ans:
C
Solution: The normal force is mgcosθ. For a horizontal surface, FN = mg. At any angle FN < mg and Ff is proportional to FN.
32. Question
The graph at left shows the relationship between the mass of a number of rubber stoppers and their resulting weight on some far-off planet. The slope of the graph is a representation of the:
(A) mass of a stopper
(B) density of a stopper
(C) acceleration due to gravity
(D) number of stoppers for each unit of weight
Answer/Explanation
Ans:
C
Solution: Slope = Δy/Δx = Weight/mass = acceleration due to gravity
33. Question
Two masses, m1 and m2, are connected by a cord and arranged as shown in the diagram with m1 sliding along on a frictionless surface and m2 hanging from a light frictionless pulley. What would be the mass of the falling mass, m2, if both the sliding mass, m1, and the tension, T, in the cord were known?
(A) \(\frac{m_{1}g – T}{g}\) (B) \(\frac{1}{2}Tg\) (C) \(\frac{m_{1}\left ( T-g \right )}{\left ( gm_{1}-T \right )}\) (D) \(\frac{Tm_{1}}{\left ( gm_{1}-T \right )}\)
Answer/Explanation
Ans:
D
Solution: Newton’s second law applied to m1: T = m1a, or a = T/m1, substitute this into Newton’s second law for the hanging mass: m2g – T = m2a
34. Question
A mass is suspended from the roof of a lift (elevator) by means of a spring balance. The lift (elevator) is moving upwards and the readings of the spring balance are noted as follows:
Speeding up: RU Constant speed: RC Slowing down:RD
Which of the following is a correct relationship between the readings?
(A) RU > RC (B) RU = RD (C) RC < RD (D) RC < RD
Answer/Explanation
Ans:
A
Solution: Elevator physics: R represents the scale reading. \(\sum \) F = ma; R – mg = ma, or R = m(g + a). This ranks the value of R from largest to smallest as accelerating upward, constant speed, accelerating downward
35. Question
A small box of mass m is placed on top of a larger box of mass 2m as shown in the diagram at right. When a force F is applied to the large box, both boxes accelerate to the right with the same acceleration. If the coefficient of friction between all surfaces is μ, what would be the force accelerating the smaller mass?
(A) \(\frac{F}{3}- mg\mu \) (B) F – 3mgμ (C) F – mgμ (D) \(\frac{F-mg\mu }{3}\)
Answer/Explanation
Ans:
A
Solution: \(\sum \) F = ma for the whole system gives F – μ(3m)g = (3m)a and solving for a gives a = (F –3μmg)/3m. For the top block, Fm = ma = m[(F – 3μmg)/3m]
36. Question
A 6.0 kg block initially at rest is pushed against a wall by a 100 N force as shown. The coefficient of kinetic friction is 0.30 while the coefficient of static friction is 0.50. What is true of the friction acting on the block after a time of 1 second?
(A) Static friction acts upward on the block.
(B) Kinetic friction acts upward on the block
(C) Kinetic friction acts downward on the block.
(D) Static friction acts downward on the block.
Answer/Explanation
Ans:
C
Solution: The normal force comes from the perpendicular component of the applied force which is Fcosθ = 50 N. The maximum value of static friction is then μFN = 25 N. The upward component of the applied force is Fsinθ = 87 N. \(\sum \) Fy = Fup – mg = 87 N– 60 N > 25 N. Since the net force on the block is great than static friction can hold, the block will begin moving up the wall. Since it is in motion, kinetic friction is acting opposite the direction of the block’s motion
37. Question
A homeowner pushes a lawn mower across a horizontal patch of grass with a constant speed by applying a force P. The arrows in the diagram correctly indicate the directions but not necessarily the magnitudes of the various forces on the lawn mower. Which of the following relations among the various force magnitudes, W, f, N, P is correct?
(A) P > f and N > W
(B) P < f and N = W
(C) P > f and N < W
(D) P = f and N > W
Answer/Explanation
Ans:
A
Solution: Since P is at a downward angle, the normal force is increased. Since a component of P balances the frictional force, P itself must be larger than f.
38. Question
A mass, M, is at rest on a frictionless surface, connected to an ideal horizontal spring that is unstretched. A person extends the spring 30 cm from equilibrium and holds it at this location by applying a 10 N force. The spring is brought back to equilibrium and the mass connected to it is now doubled to 2M. If the spring is extended back 30 cm from equilibrium, what is the necessary force applied by the person to hold the mass stationary there?
(A) 20.0 N (B) 14.1 N (C) 10.0 N (D) 7.07 N
Answer/Explanation
Ans:
C
Solution: Since the force is applied horizontally, the mass has no effect.
39. Question
A crate of toys remains at rest on a sleigh as the sleigh is pulled up a hill with an increasing speed. The crate is not fastened down to the sleigh. What force is responsible for the crate’s increase in speed up the hill?
(A) the contact force (normal force) of the ground on the sleigh
(B) the force of static friction of the sleigh on the crate
(C) the gravitational force acting on the sleigh
(D) no force is needed
Answer/Explanation
Ans:
B
Solution: The only force in the direction of the crate’s acceleration is the force of friction from the sleigh
40. Question
A box slides to the right across a horizontal floor. A person called Ted exerts a force T to the right on the box. A person called Mario exerts a force M to the left, which is half as large as the force T. Given that there is friction f and the box accelerates to the right, rank the sizes of these three forces exerted on the box.
(A) f < M < T (B) M < f < T (C) M < T < f (D) f = M < T
Answer/Explanation
Ans:
A
Solution: Given that the box accelerates toward Ted, Ted’s force must be greater than Mario’s force plus the force of friction. Since Mario’s force is ½ of Ted’s force, the force of frction must be less than half of Ted’s force.
41. Question
A spaceman of mass 80 kg is sitting in a spacecraft near the surface of the Earth. The spacecraft is accelerating upward at five times the acceleration due to gravity. What is the force of the spaceman on the spacecraft?
(A) 4800 N (B) 4000 N (C) 3200 N (D) 800 N
Answer/Explanation
Ans:
A
Solution: \(\sum \) F = ma; F – mg = m(5g) or F = 6mg
42. Question
Two identical blocks of weight W are placed one on top of the other as shown in the diagram above. The upper block is tied to the wall. The lower block is pulled to the right with a force F. The coefficient of static friction between all surfaces in contact is μ. What is the largest force F that can be exerted before the lower block starts to slip?
(A) μW (B) 2μW (C) 3μW (D) 3μW/2
Answer/Explanation
Ans:
C
Solution: Between the lower block and the tabletop, there is a force of friction to the left of maximum magnitude μ(2W) as both blocks are pushing down on the tabletop. There is also a force of friction acting to the left on the upper surface of the lower block due to the upper block of maximum magnitude μW. The total maximum static frictional force holding the lower block in place is therefore μ(2W) + μ
43. Question
A force F is used to hold a block of mass m on an incline as shown in the diagram (see above). The plane makes an angle of θ with the horizontal and F is perpendicular to the plane. The coefficient of friction between the plane and the block is μ. What is the minimum force, F, necessary to keep the block at rest?
(A) mgcosθ (B) mgsinθ (C) mgsinθ/μ (D) mg(sinθ – μcosθ)/μ
Answer/Explanation
Ans:
D
Solution: The normal force on the block can be found from \(\sum \) Fy = 0 = FN – mgcosθ – F. The force of friction necessary to hold the block in place is mgsinθ. Setting the force of friction equal to mgsinθ gives μFN = mgsinθ = μ(F + mgcosθ)
44. Question
When the speed of a rear-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
(A) backward on the front tires and forward on the rear tires
(B) forward on the front tires and backward on the rear tires
(C) forward on all tires
(D) backward on all tires
Answer/Explanation
Ans:
A
Solution: This is a tricky one. In order to move the car forward, the rear tires roll back against the ground, the force of friction pushing forward on the rear tires. The front tires, however, are not trying to roll on their own, rather they begin rolling due to the friction acting backward, increasing their rate of rotation
45. Question
Given the three masses as shown in the diagram above, if the coefficient of kinetic friction between the large mass (m2) and the table is μ, what would be the upward acceleration of the small mass (m3)? The mass and friction of the cords and pulleys are small enough to produce a negligible effect on the system.
(A) g(m1 + m2μ)/(m1 + m2 + m3) (B) gμ(m1 + m2 + m3)/ (m1 – m2 – m3)
(C) gμ(m1 – m2 – m3)/ (m1 + m2 + m3) (D) g(m1 – μm2 – m3)/ (m1 + m2 + m3)
Answer/Explanation
Ans:
D
Solution: The external forces acting on the system of masses are the weights of block 1 (pulling the system to the left), the weight of block 3 (pulling the system to the right) and the force of friction on block 2 (pulling the system to the left with a magnitude μFN = μm2g) \(\sum \) Fexternal = mtotala gives (m1g – μm2g – m3g) = (m1 + m2 + m3)a
Questions 47-48
Three identical laboratory carts A, B, and C are each subject to a constant force FA, FB, and FC, respectively. One or more of these forces may be zero. The diagram below shows the position of each cart at each second of an 8.0 second interval.
47. Question
Which car has the greatest average velocity during the interval?
(A) A (B) B (C) C (D) all three average velocities are equal
Answer/Explanation
Ans:
D
Solution: As they are all at the same position after 8 seconds, they all have the same average velocity
48. Question
How does the magnitude of the force acting on each car compare?
(A) FA > FB > FC (B) FA = FC > FB (C) FA > FC = FB (D) FA = FB > FC
Answer/Explanation
Ans:
B
Solution: Car A decelerates with the same magnitude that C accelerates. Car B is moving at constant speed, which means FB = 0.
49. Question
A skydiver is falling at terminal velocity before opening her parachute. After opening her parachute, she falls at a much smaller terminal velocity. How does the total upward force before she opens her parachute compare to the total upward force after she opens her parachute?
(A) The ratio of the forces is equal to the ratio of the velocities.
(B) The upward force with the parachute will depend on the size of the parachute.
(C) The upward force before the parachute will be greater because of the greater velocity.
(D) The upward force in both cases must be the same.
Answer/Explanation
Ans:
D
Solution: When falling with terminal velocity, the force of air resistance equals your weight, regardless of the speed.
50. Question
Each of the diagrams below represents two weights connected by a massless string which passes over a massless, frictionless pulley. In which diagram will the magnitude of the acceleration be the largest?
Answer/Explanation
Ans:
A
Solution: For each case, \(\sum \) Fexternal = mtotala gives Mg – mg = (M + m)a, or a = \(\frac{M-m}{M+m}g\)
51. Question
A simple Atwood’s machine is shown in the diagram above. It is composed of a frictionless lightweight pulley with two cubes connected by a light string. If cube A has a mass of 4.0 kg and cube B has a mass of 6.0 kg, the system will move such that cube B accelerates downwards. What would be the tension in the two parts of the string between the pulley and the cubes?
(A) TA = 47 N ; TB = 71 N (B) TA = 47 N ; TB = 47 N (C) TA = 47 N ; TB = 42 N
(D) TA = 39 N ; TB = 39 N
Answer/Explanation
Ans:
B
Solution: The two ends of the light string must have the same tension, eliminating choices A and C. If choice D was correct, both masses would be accelerating downward and TA must be greater than the weight of block A.
52. Question
A simple Atwood’s machine remains motionless when equal masses M are placed on each end of the chord. When a small mass m is added to one side, the masses have an acceleration a. What is M? You may neglect friction and the mass of the cord and pulley
(A) \(\frac{m(g-a)}{2a}\) (B) \(\frac{2m(g-a)}{a}\) (C) \(\frac{2m(g+a)}{a}\) (D) \(\frac{m(g+a)}{2a}\)
Answer/Explanation
Ans:
A
Solution: \(\sum \) Fexternal = mtotala gives (M + m)g – Mg = (2M + m)a
53. Question
Block 1 is stacked on top of block 2. Block 2 is connected by a light cord to block 3, which is pulled along a frictionless surface with a force F as shown in the diagram. Block 1 is accelerated at the same rate as block 2 because of the frictional forces between the two blocks. If all three blocks have the same mass m, what is the minimum coefficient of static friction between block 1 and block 2?
(A) 2F/mg (B) F/mg (C) 3F/2mg (D) F/3mg
Answer/Explanation
Ans:
D
Solution: As the entire system moves as one, F = (3m)a, or a = F/(3m). The force of friction acting on block 1 is the force moving block 1 and we have μmg = m(F/(3m))
54. Question
Three blocks (m1, m2, and m3) are sliding at a constant velocity across a rough surface as shown in the diagram above. The coefficient of kinetic friction between each block and the surface is μ. What would be the force of m1 on m2?
(A) (m2 + m3)gμ (B) F – (m2 – m3)gμ (C) F (D) m1gμ – (m2 + m3)gμ
Answer/Explanation
Ans:
A
Solution: Since the system is moving at constant velocity, m1 is pushing m2 and m3 with a force equal to the force of friction acting on those two blocks, which is μ(FN2 + FN3)