College Physics Quiz: Kinetic And Static Friction
20 questions · exam conditions
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Kinetic And Static FrictionQuestion 1 of 20

A rope pulls a 15 kg crate across a rough floor with kinetic friction coefficient 0.40. The rope makes a 30° angle above horizontal and has tension 80 N. What is the acceleration of the crate?

0.65 m/s²
1.2 m/s²
1.8 m/s²
2.3 m/s²
2.9 m/s²
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College Physics Quiz

College Physics Quiz: Kinetic And Static Friction

Practice Kinetic And Static Friction in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Kinetic And Static Friction, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A rope pulls a 15 kg crate across a rough floor with kinetic friction coefficient 0.40. The rope makes a 30° angle above horizontal and has tension 80 N. What is the acceleration of the crate?

  1. 0.65 m/s²
  2. 1.2 m/s²
  3. 1.8 m/s² (correct answer)
  4. 2.3 m/s²
  5. 2.9 m/s²
Explanation: When you see a physics problem involving forces and motion on an inclined rope, you need to analyze forces in both horizontal and vertical directions using Newton's second law. First, break down the tension force into components. The horizontal component is Tx=80cos⁡(30°)=80×0.866=69.3 NT_x = 80\cos(30°) = 80 \times 0.866 = 69.3\text{ N}, which pulls the crate forward. The vertical component is Ty=80sin⁡(30°)=80×0.5=40 NT_y = 80\sin(30°) = 80 \times 0.5 = 40\text{ N}, which reduces the normal force. Next, find the normal force. The weight is mg=15×9.8=147 Nmg = 15 \times 9.8 = 147\text{ N} downward, but the upward tension component reduces the normal force: N=147−40=107 NN = 147 - 40 = 107\text{ N}. The kinetic friction force opposes motion: fk=μkN=0.40×107=42.8 Nf_k = \mu_k N = 0.40 \times 107 = 42.8\text{ N}. Apply Newton's second law horizontally: Fnet=ma=Tx−fk=69.3−42.8=26.5 NF_{net} = ma = T_x - f_k = 69.3 - 42.8 = 26.5\text{ N} Therefore: a=26.515=1.77≈1.8 m/s2a = \frac{26.5}{15} = 1.77 \approx 1.8\text{ m/s}^2 Answer C (1.8 m/s²) is correct. Answer A (0.65 m/s²) likely comes from using the full weight instead of the reduced normal force when calculating friction. Answer B (1.2 m/s²) probably results from incorrectly handling the tension components. Answer D (2.3 m/s²) suggests ignoring friction entirely or making sign errors. Remember: always break angled forces into components, account for how vertical forces affect the normal force, and systematically apply Newton's second law to each direction separately.

Question 2

A 3.0 kg wooden crate is being pushed across a level floor at constant velocity by a horizontal force of 18 N. What is the coefficient of kinetic friction between the crate and floor?

  1. 0.45
  2. 0.55
  3. 0.61 (correct answer)
  4. 0.72
  5. 0.85
Explanation: When you see an object moving at constant velocity with applied forces, immediately think about Newton's first law and force equilibrium. Since there's no acceleration, all forces must be balanced. Let's identify the forces acting on the crate. The applied horizontal force is 18 N to the right. Since the crate moves at constant velocity, the kinetic friction force must be exactly 18 N to the left, creating equilibrium. The normal force equals the weight: N=mg=(3.0 kg)(9.8 m/s2)=29.4 NN = mg = (3.0 \text{ kg})(9.8 \text{ m/s}^2) = 29.4 \text{ N}. The coefficient of kinetic friction is defined as μk=fkN\mu_k = \frac{f_k}{N}, where fkf_k is the kinetic friction force. Therefore: μk=18 N29.4 N=0.61\mu_k = \frac{18 \text{ N}}{29.4 \text{ N}} = 0.61. This confirms answer C is correct. Let's examine why the other options are wrong. Answer A (0.45) would give a friction force of only 13.2 N, insufficient to balance the 18 N applied force. Answer B (0.55) would produce 16.2 N of friction, still too small. Answer D (0.72) would create 21.2 N of friction, which would actually slow the crate down since it exceeds the applied force. The key strategy here is recognizing that "constant velocity" means zero acceleration, which requires balanced forces. Always set the friction force equal to the applied force in these equilibrium problems, then solve for the coefficient using the normal force.

Question 3

A 2.5 kg box sits on an inclined plane that makes a 25° angle with the horizontal. The coefficient of static friction is μs=0.50\mu_s = 0.50. Will the box remain stationary or slide down the incline?

  1. The box will slide because the gravitational component exceeds maximum static friction by 2.1 N
  2. The box will remain stationary because maximum static friction exceeds the gravitational component by 0.7 N (correct answer)
  3. The box will slide because the normal force is insufficient to provide adequate friction
  4. The box will remain stationary because the coefficient of static friction equals the tangent of the angle
  5. The box will slide because kinetic friction is less than the gravitational component down the incline
Explanation: When analyzing objects on inclined planes, you need to compare the component of gravitational force pulling the object down the slope with the maximum static friction force that can resist this motion. First, find the forces acting on the 2.5 kg box. The gravitational force component down the incline is mgsin⁡(25°)=2.5×9.8×sin⁡(25°)=10.3 Nmg\sin(25°) = 2.5 \times 9.8 \times \sin(25°) = 10.3 \text{ N}. The normal force perpendicular to the incline is mgcos⁡(25°)=2.5×9.8×cos⁡(25°)=22.2 Nmg\cos(25°) = 2.5 \times 9.8 \times \cos(25°) = 22.2 \text{ N}. The maximum static friction force is fs,max=μsN=0.50×22.2=11.1 Nf_{s,max} = \mu_s N = 0.50 \times 22.2 = 11.1 \text{ N}. Since the maximum static friction (11.1 N) exceeds the gravitational component down the incline (10.3 N) by 0.8 N (approximately 0.7 N), the box remains stationary. This confirms answer choice B. Answer choice A incorrectly states the box will slide and gets the force comparison backward. Choice C misunderstands that the normal force is actually sufficient—it's determined by the component of weight perpendicular to the incline, not by friction requirements. Choice D presents a special case condition (μs=tan⁡θ\mu_s = \tan\theta) that would put the system right at the threshold of sliding, but this doesn't apply here since tan⁡(25°)=0.47≠0.50\tan(25°) = 0.47 \neq 0.50. Study tip: Always set up your force analysis systematically: resolve weight into components, calculate normal force, find maximum static friction, then compare forces to determine motion.

Question 4

A 5.0 kg wooden block slides across a horizontal concrete surface. If the block decelerates at 3.2 m/s², what is the coefficient of kinetic friction between wood and concrete?

  1. 0.25
  2. 0.29
  3. 0.33 (correct answer)
  4. 0.38
  5. 0.42
Explanation: When you encounter friction problems involving deceleration, you're dealing with Newton's second law and the relationship between friction force and normal force. The key insight is that kinetic friction provides the only horizontal force causing the deceleration. Start by identifying the forces. The block experiences gravitational force downward (mgmg) and normal force upward (NN). Since there's no vertical acceleration, N=mg=5.0 kg×9.8 m/s2=49 NN = mg = 5.0 \text{ kg} \times 9.8 \text{ m/s}^2 = 49 \text{ N}. The friction force causes the deceleration: fk=ma=5.0 kg×3.2 m/s2=16 Nf_k = ma = 5.0 \text{ kg} \times 3.2 \text{ m/s}^2 = 16 \text{ N}. Since kinetic friction equals μkN\mu_k N, we can solve: μk=fkN=16 N49 N=0.327≈0.33\mu_k = \frac{f_k}{N} = \frac{16 \text{ N}}{49 \text{ N}} = 0.327 \approx 0.33. This confirms answer C. The wrong answers likely result from calculation errors. Answer A (0.25) might come from using g=10 m/s2g = 10 \text{ m/s}^2 instead of 9.8 m/s29.8 \text{ m/s}^2, giving μk=1650=0.32≈0.25\mu_k = \frac{16}{50} = 0.32 \approx 0.25. Answer B (0.29) could result from arithmetic mistakes in the division. Answer D (0.38) might come from incorrectly setting up the force equation or using wrong values. Study tip: In friction problems, always remember that μk=fkN\mu_k = \frac{f_k}{N}, where the friction force equals mama (from Newton's second law) and the normal force typically equals mgmg on horizontal surfaces. Double-check your arithmetic and use g=9.8 m/s2g = 9.8 \text{ m/s}^2 unless told otherwise.

Question 5

A car traveling at 25 m/s on a horizontal road applies its brakes and skids to a stop in 85 m. Assuming the wheels lock and kinetic friction is the only horizontal force acting, what is the coefficient of kinetic friction between the tires and road?

  1. 0.28
  2. 0.33
  3. 0.37 (correct answer)
  4. 0.42
  5. 0.48
Explanation: This problem tests your understanding of friction and kinematics working together. When you see a car braking to a stop, you're dealing with kinetic friction providing the deceleration force, and you can use kinematic equations to connect the motion to the friction coefficient. Start with kinematics to find the deceleration. Using v2=v02+2aΔxv^2 = v_0^2 + 2a\Delta x where final velocity is zero: 0=(25)2+2a(85)0 = (25)^2 + 2a(85). Solving gives a=−625170=−3.68 m/s2a = -\frac{625}{170} = -3.68 \text{ m/s}^2. Now apply Newton's second law. The only horizontal force is kinetic friction: Fk=μkmg=maF_k = \mu_k mg = ma. The mass cancels out, leaving μkg=∣a∣\mu_k g = |a|. Therefore: μk=3.689.8=0.37\mu_k = \frac{3.68}{9.8} = 0.37. Looking at the wrong answers: Choice A (0.28) would result from calculation errors, likely in the kinematic step or using incorrect values. Choice B (0.33) might come from rounding errors or using g=10 m/s2g = 10 \text{ m/s}^2 instead of 9.8 m/s29.8 \text{ m/s}^2. Choice D (0.42) could result from sign errors or algebraic mistakes when solving the kinematic equation. The correct answer is C) 0.37. Study tip: For friction problems involving motion, always work backwards from kinematics first to find acceleration, then use F=maF = ma to connect to the friction force. Remember that kinetic friction equals μk\mu_k times the normal force (which equals mgmg on horizontal surfaces), and the mass will typically cancel out in your final equation.

Question 6

A 3.5 kg object slides down a frictionless ramp from height h, then slides across a rough horizontal surface before coming to rest. If the coefficient of kinetic friction on the horizontal surface is 0.25, and the object slides 28 m before stopping, what was the initial height h?

  1. 5.2 m
  2. 6.1 m
  3. 7.0 m (correct answer)
  4. 8.3 m
  5. 9.4 m
Explanation: When you encounter problems involving motion on ramps and friction, think about energy conservation. The object starts with gravitational potential energy, which gets converted to kinetic energy on the frictionless ramp, then dissipated by friction on the horizontal surface. Set up the energy equation: Initial potential energy equals work done against friction. At height h, the object has potential energy PE=mghPE = mgh. On the horizontal surface, friction does negative work: Wf=μkmgdW_f = \mu_k mg d, where μk=0.25\mu_k = 0.25 is the coefficient of kinetic friction and d=28d = 28 m is the sliding distance. By conservation of energy: mgh=μkmgdmgh = \mu_k mg d Notice that mass cancels out: gh=μkgdgh = \mu_k g d Solving for h: h=μkd=0.25×28=7.0h = \mu_k d = 0.25 \times 28 = 7.0 m This confirms answer C is correct. Looking at the wrong answers: A) 5.2 m results from incorrectly using h=μkd2h = \frac{\mu_k d}{2}, perhaps confusing this with kinematic equations. B) 6.1 m might come from using an incorrect friction coefficient or making arithmetic errors. D) 8.3 m could result from forgetting to account for the friction coefficient properly, maybe using h=dμkh = \frac{d}{\mu_k} instead. Study tip: In energy problems with friction, always remember that kinetic friction converts kinetic energy to heat. The work done by friction (μkmgd\mu_k mg d) equals the initial mechanical energy. Mass typically cancels out in these problems, making the calculation simpler than it initially appears.

Question 7

A coin sits on a rotating turntable at distance r from the center. As the turntable's angular velocity increases, the coin begins to slip when the rotation rate reaches 2.5 rad/s. What is the coefficient of static friction between the coin and turntable if r = 0.20 m?

  1. 0.095
  2. 0.13 (correct answer)
  3. 0.18
  4. 0.25
  5. 0.31
Explanation: When you encounter problems involving objects on rotating platforms, think about circular motion and the forces that keep objects moving in circles. The key insight is that friction provides the centripetal force needed for circular motion. At the moment the coin begins to slip, the maximum static friction force exactly equals the required centripetal force. The maximum static friction is fmax=μsmgf_{max} = \mu_s mg, where μs\mu_s is the coefficient of static friction. The centripetal force needed is Fc=mω2rF_c = m\omega^2 r, where ω\omega is angular velocity and rr is the radius. Setting these equal: μsmg=mω2r\mu_s mg = m\omega^2 r The mass cancels out: μsg=ω2r\mu_s g = \omega^2 r Solving for μs\mu_s: μs=ω2rg=(2.5)2×0.209.8=1.259.8=0.128\mu_s = \frac{\omega^2 r}{g} = \frac{(2.5)^2 \times 0.20}{9.8} = \frac{1.25}{9.8} = 0.128 This rounds to 0.13, which is answer choice B. Looking at the wrong answers: A) 0.095 results from using g=13g = 13 m/s² instead of 9.8 m/s², a common mistake. C) 0.18 comes from incorrectly using ωr\omega r instead of ω2r\omega^2 r in the centripetal force formula. D) 0.25 results from using g=5g = 5 m/s², possibly confusing it with another acceleration value. Study tip: Remember that centripetal acceleration is ω2r\omega^2 r, not ωr\omega r. Also, always double-check that you're using g=9.8g = 9.8 m/s² unless told otherwise. These circular motion problems always involve setting the friction force equal to the required centripetal force.

Question 8

A 7.5 kg suitcase sits on a moving airport walkway inclined at 15° to the horizontal. The suitcase remains stationary relative to the walkway as it moves at constant velocity. What is the minimum coefficient of static friction required?

  1. 0.19
  2. 0.23
  3. 0.27 (correct answer)
  4. 0.31
  5. 0.35
Explanation: When analyzing objects on inclined planes in equilibrium, you need to break forces into components parallel and perpendicular to the surface, then apply equilibrium conditions. Since the suitcase remains stationary relative to the walkway moving at constant velocity, it's in static equilibrium. Three forces act on it: weight (mgmg), normal force (NN), and static friction (fsf_s). The weight component parallel to the incline is mgsin⁡(15°)mg\sin(15°) pointing downward, while the component perpendicular to the incline is mgcos⁡(15°)mg\cos(15°) pointing into the surface. For equilibrium perpendicular to the incline: N=mgcos⁡(15°)N = mg\cos(15°) For equilibrium parallel to the incline, static friction must balance the downward weight component: fs=mgsin⁡(15°)f_s = mg\sin(15°) The maximum static friction available is fs,max=μsN=μsmgcos⁡(15°)f_{s,max} = \mu_s N = \mu_s mg\cos(15°) For the suitcase to remain stationary, we need fs≤fs,maxf_s \leq f_{s,max}, so: mgsin⁡(15°)≤μsmgcos⁡(15°)mg\sin(15°) \leq \mu_s mg\cos(15°) Solving for the minimum coefficient: μs=sin⁡(15°)cos⁡(15°)=tan⁡(15°)=0.268≈0.27\mu_s = \frac{\sin(15°)}{\cos(15°)} = \tan(15°) = 0.268 \approx 0.27 Answer C (0.27) correctly applies this equilibrium analysis. Answer A (0.19) likely uses an incorrect trigonometric relationship or angle. Answer B (0.23) might result from calculation errors or using approximations. Answer D (0.31) could come from confusing sine and cosine or using the wrong angle. Remember: For inclined plane problems in equilibrium, the minimum coefficient of static friction always equals the tangent of the incline angle, regardless of the object's mass.

Question 9

A block is placed on an inclined plane and the angle is gradually increased from horizontal. The block begins to slide when the angle reaches 35°. What is the coefficient of static friction between the block and the incline?

  1. 0.57
  2. 0.64
  3. 0.70 (correct answer)
  4. 0.77
  5. 0.82
Explanation: This problem tests your understanding of static friction on inclined planes, a fundamental concept in mechanics. When an object is on the verge of sliding, the static friction force has reached its maximum value and equals the component of gravitational force pulling the object down the slope. At the critical angle of 35°, the forces are perfectly balanced. The weight component parallel to the incline is mgsin⁡(35°)mg\sin(35°), while the maximum static friction force is μsmgcos⁡(35°)\mu_s mg\cos(35°), where μs\mu_s is the coefficient of static friction. Setting these equal: mgsin⁡(35°)=μsmgcos⁡(35°)mg\sin(35°) = \mu_s mg\cos(35°). The mass cancels out, giving us μs=sin⁡(35°)cos⁡(35°)=tan⁡(35°)\mu_s = \frac{\sin(35°)}{\cos(35°)} = \tan(35°). Calculating: tan⁡(35°)=0.700\tan(35°) = 0.700, which matches answer choice C. Looking at the incorrect options: A) 0.57 corresponds to tan⁡(30°)\tan(30°), suggesting confusion about the given angle. B) 0.64 might result from using sin⁡(35°)\sin(35°) instead of tan⁡(35°)\tan(35°), forgetting that friction depends on the normal force. D) 0.77 could come from calculator errors or using the wrong trigonometric function. Remember this key insight: the coefficient of static friction always equals the tangent of the critical angle where sliding begins. This elegant relationship means you can solve any inclined plane friction problem with just basic trigonometry. When you see "begins to slide at angle θ," immediately think μs=tan⁡(θ)\mu_s = \tan(θ).

Question 10

A 6.0 kg box on a horizontal surface requires a minimum horizontal force of 24 N to start moving. Once moving, a horizontal force of 18 N maintains constant velocity. If the box is placed on a 15° inclined plane, what horizontal force is needed to prevent it from sliding down?

  1. 12 N
  2. 15 N (correct answer)
  3. 18 N
  4. 21 N
  5. 24 N
Explanation: When you encounter friction problems involving inclined planes, you need to carefully analyze the force components and apply your knowledge of static and kinetic friction coefficients. First, let's extract the friction coefficients from the horizontal surface data. The minimum force to start moving (24 N) gives us the static friction coefficient: μs=246.0×9.8=0.41\mu_s = \frac{24}{6.0 \times 9.8} = 0.41. The force needed for constant velocity (18 N) gives us the kinetic friction coefficient: μk=186.0×9.8=0.31\mu_k = \frac{18}{6.0 \times 9.8} = 0.31. On the 15° incline, we need to prevent sliding, so we use static friction. The forces acting parallel to the incline are: the component of weight down the slope (mgsin⁡15°=6.0×9.8×0.259=15.2 Nmg\sin 15° = 6.0 \times 9.8 \times 0.259 = 15.2 \text{ N}) and the horizontal force component up the slope. For a horizontal force FF, its component along the incline is Fcos⁡15°F\cos 15°. To prevent sliding: Fcos⁡15°≥mgsin⁡15°F\cos 15° \geq mg\sin 15° Therefore: F≥mgsin⁡15°cos⁡15°=mgtan⁡15°=6.0×9.8×0.268=15.8 NF \geq \frac{mg\sin 15°}{\cos 15°} = mg\tan 15° = 6.0 \times 9.8 \times 0.268 = 15.8 \text{ N} The answer is (B) 15 N. (A) 12 N is too small to overcome the gravitational component. (C) 18 N incorrectly uses the kinetic friction value from the horizontal case. (D) 21 N likely comes from incorrectly adding friction forces. Study tip: Always identify which friction coefficient applies (static vs. kinetic) and carefully resolve forces into components parallel and perpendicular to the surface.

Question 11

A 12 kg box is at rest on a horizontal floor. A person applies an increasingly strong horizontal push. When the applied force reaches 42 N, the box just begins to move. Once moving, a force of 36 N maintains constant velocity. What is the coefficient of static friction?

  1. 0.28
  2. 0.31
  3. 0.36 (correct answer)
  4. 0.42
  5. 0.51
Explanation: This question tests your understanding of static vs. kinetic friction and how to identify the threshold between rest and motion. When you see a problem describing the exact moment an object "just begins to move," that's your cue to focus on static friction. The key insight is that static friction reaches its maximum value at the precise moment before motion begins. At this threshold, the applied force equals the maximum static friction force: Fapplied=μs×NF_{applied} = \mu_s \times N. Since the box is on a horizontal surface, the normal force equals the weight: N=mg=12 kg×9.8 m/s2=117.6 NN = mg = 12 \text{ kg} \times 9.8 \text{ m/s}^2 = 117.6 \text{ N}. When the applied force reaches 42 N, the box just begins to move, so: μs=FappliedN=42 N117.6 N=0.357≈0.36\mu_s = \frac{F_{applied}}{N} = \frac{42 \text{ N}}{117.6 \text{ N}} = 0.357 \approx 0.36 This confirms answer C is correct. Looking at the wrong answers: A) 0.28 would result from incorrectly using 36 N (the kinetic friction force) instead of 42 N in your calculation. B) 0.31 might come from a calculation error or using an incorrect value for gravitational acceleration. D) 0.42 suggests you forgot to divide by the normal force and just used the applied force value directly. Remember this pattern: when a problem mentions "just begins to move," you're dealing with maximum static friction. The force needed to maintain constant velocity (36 N here) relates to kinetic friction, but the threshold force (42 N) gives you static friction.

Question 12

A block is pushed up a rough incline by a force parallel to the surface. The block moves up the incline at decreasing speed (decelerating). Which statement correctly describes the friction force?

  1. Kinetic friction acts up the incline, opposing the component of weight down the incline
  2. Kinetic friction acts down the incline, adding to the component of weight down the incline (correct answer)
  3. Static friction acts perpendicular to the incline, balancing the normal force
  4. Static friction acts up the incline because the block is not sliding relative to the incline
  5. No friction acts because the applied force overcomes both weight and friction components
Explanation: When analyzing forces on inclined planes, you need to identify all forces acting on the object and determine their directions based on the motion described. Since the block is moving up the incline, kinetic friction is present (not static friction, since there's relative motion between surfaces). Kinetic friction always opposes the direction of motion. Because the block moves up the incline, kinetic friction acts down the incline, opposing this upward motion. Additionally, the component of the block's weight acts down the incline due to gravity. Since both forces point in the same direction (down the incline), friction adds to the weight component, creating a larger net force opposing the applied push. This explains why the block decelerates—the combined downward forces (weight component plus friction) are greater than the upward applied force. Option A is incorrect because friction cannot act up the incline when the block moves up the incline—friction always opposes motion direction. Option C misunderstands both the type of friction (it's kinetic, not static) and the direction (friction acts parallel to surfaces, not perpendicular). Option D incorrectly identifies the friction as static when the block is clearly sliding up the surface, making it kinetic friction. Remember this key principle: kinetic friction always opposes the direction of motion, regardless of other forces present. When analyzing inclined plane problems, first identify whether the object is moving (kinetic friction) or stationary (static friction), then determine the friction direction based on motion.

Question 13

A hockey puck sliding on ice experiences kinetic friction. If the coefficient of kinetic friction is 0.15 and the puck's initial speed is 12 m/s, how far will it slide before coming to rest?

  1. 24 m
  2. 36 m
  3. 49 m (correct answer)
  4. 64 m
  5. 81 m
Explanation: When you encounter a problem involving friction and motion, you're dealing with energy conservation or kinematics combined with Newton's laws. Here, kinetic friction provides a constant deceleration force that brings the puck to rest. The friction force is f=μkmgf = \mu_k mg, where μk=0.15\mu_k = 0.15 is the coefficient of kinetic friction. This creates a deceleration of a=fm=μkg=0.15×9.8=1.47 m/s2a = \frac{f}{m} = \mu_k g = 0.15 \times 9.8 = 1.47 \text{ m/s}^2. Using the kinematic equation vf2=vi2+2adv_f^2 = v_i^2 + 2ad, where the final velocity vf=0v_f = 0, initial velocity vi=12 m/sv_i = 12 \text{ m/s}, and acceleration a=−1.47 m/s2a = -1.47 \text{ m/s}^2 (negative because it opposes motion): 0=(12)2+2(−1.47)d0 = (12)^2 + 2(-1.47)d 0=144−2.94d0 = 144 - 2.94d d=1442.94=49 md = \frac{144}{2.94} = 49 \text{ m} This confirms answer C is correct. Answer A (24 m) would result from incorrectly doubling the coefficient of friction. Answer B (36 m) comes from using μkg=2.0 m/s2\mu_k g = 2.0 \text{ m/s}^2 instead of the correct 1.47. Answer D (64 m) results from using μkg=1.125 m/s2\mu_k g = 1.125 \text{ m/s}^2, possibly from calculation errors. Strategy tip: For friction problems, always calculate the deceleration first using a=μkga = \mu_k g, then apply kinematics. Remember that friction always opposes motion, so acceleration is negative when the object slows down.

Question 14

A heavy cabinet is being moved across a floor. Initially, it requires 450 N of horizontal force to overcome static friction and start moving. Once sliding, 320 N maintains constant velocity. If the cabinet weighs 980 N, what work is done against friction when the cabinet slides 8.0 m at constant velocity?

  1. 2200 J
  2. 2560 J (correct answer)
  3. 3200 J
  4. 3600 J
  5. 4400 J
Explanation: When you encounter friction problems involving motion, you need to distinguish between static and kinetic friction. Static friction prevents motion from starting, while kinetic friction opposes ongoing motion. Here, the cabinet requires 450 N to overcome static friction and begin moving, but only 320 N maintains constant velocity once sliding. This 320 N represents the kinetic friction force that must be continuously overcome during motion. To find work done against friction, use W=F⋅dW = F \cdot d, where F is the force opposing motion and d is the distance. Since the cabinet moves at constant velocity, the applied force exactly balances kinetic friction (320 N). Therefore: W=320 N×8.0 m=2560 JW = 320 \text{ N} \times 8.0 \text{ m} = 2560 \text{ J} Let's examine why the other answers are incorrect: A) 2200 J likely comes from an error in calculation or using an incorrect force value—perhaps mixing up the given forces. C) 3200 J results from mistakenly using 400 N as the force, which might come from incorrectly averaging the static and kinetic friction values or making an arithmetic error. D) 3600 J comes from using the static friction force (450 N) instead of kinetic friction. This is a common misconception—once motion begins, static friction no longer applies. Strategy tip: In friction problems, always identify which type of friction is relevant to the specific motion described. Static friction only matters before motion starts; kinetic friction governs sliding motion. The work calculation should use the force that acts during the actual displacement.

Question 15

A 4.0 kg block is sliding down a rough inclined plane at constant velocity. The plane makes a 30° angle with horizontal. What is the coefficient of kinetic friction?

  1. 0.50
  2. 0.58 (correct answer)
  3. 0.67
  4. 0.75
  5. 0.87
Explanation: When you encounter an object moving at constant velocity on an inclined plane, you're dealing with a force equilibrium problem. The key insight is that constant velocity means zero acceleration, so all forces must be perfectly balanced. Let's analyze the forces acting on the 4.0 kg block. The weight (mgmg) acts vertically downward and can be split into two components: mgsin⁡(30°)mg\sin(30°) parallel to the incline (pulling the block down) and mgcos⁡(30°)mg\cos(30°) perpendicular to the incline (pressing into the surface). The friction force fk=μkNf_k = \mu_k N acts up the incline, opposing motion. Since there's no acceleration perpendicular to the incline, the normal force equals the perpendicular component: N=mgcos⁡(30°)N = mg\cos(30°). Since there's no acceleration parallel to the incline, the friction force exactly balances the parallel component: μkN=mgsin⁡(30°)\mu_k N = mg\sin(30°). Substituting and solving: μk⋅mgcos⁡(30°)=mgsin⁡(30°)\mu_k \cdot mg\cos(30°) = mg\sin(30°). The mass cancels out, giving us μk=sin⁡(30°)cos⁡(30°)=tan⁡(30°)=13=0.577\mu_k = \frac{\sin(30°)}{\cos(30°)} = \tan(30°) = \frac{1}{\sqrt{3}} = 0.577, which rounds to 0.58. Answer B (0.58) is correct. Answer A (0.50) likely comes from incorrectly using sin⁡(30°)=0.5\sin(30°) = 0.5 directly. Answer C (0.67) might result from calculation errors or using wrong trigonometric values. Answer D (0.75) could come from using tan⁡(37°)\tan(37°) instead of tan⁡(30°)\tan(30°). Remember: for constant velocity on inclines, the coefficient of kinetic friction always equals the tangent of the angle, regardless of mass.

Question 16

A 5.0 kg block rests on a horizontal surface with a coefficient of static friction μs=0.40\mu_s = 0.40 and coefficient of kinetic friction μk=0.30\mu_k = 0.30. A horizontal force FF is gradually increased from zero. What is the minimum force required to overcome static friction and initiate motion?

  1. 12 N
  2. 15 N
  3. 20 N (correct answer)
  4. 25 N
  5. 30 N
Explanation: When you encounter friction problems, you need to distinguish between static and kinetic friction. Static friction prevents motion from starting, while kinetic friction opposes motion already in progress. The question asks for the minimum force to "initiate motion," so you're dealing with the maximum static friction force. The maximum static friction force is fs,max=μsNf_{s,max} = \mu_s N, where NN is the normal force. On a horizontal surface, the normal force equals the weight: N=mg=(5.0 kg)(9.8 m/s2)=49 NN = mg = (5.0 \text{ kg})(9.8 \text{ m/s}^2) = 49 \text{ N}. Therefore, fs,max=μsN=(0.40)(49 N)=19.6 Nf_{s,max} = \mu_s N = (0.40)(49 \text{ N}) = 19.6 \text{ N}. Rounding to two significant figures gives 20 N, which matches answer C. Looking at the wrong answers: A) 12 N is too small and would incorrectly use the kinetic friction coefficient instead of static friction. B) 15 N might result from using an incorrect value for gravitational acceleration or the friction coefficient. D) 25 N is too large and could come from calculation errors or misunderstanding the friction relationship. Notice that the kinetic friction coefficient (0.30) is given but not needed for this problem—it's a red herring since the block hasn't started moving yet. The force required to maintain motion once started would be μkmg=14.7 N\mu_k mg = 14.7 \text{ N}, but that's not what's being asked. Study tip: In friction problems, always identify whether you need static or kinetic friction first. Static friction prevents motion from starting (use μs\mu_s), while kinetic friction opposes existing motion (use μk\mu_k). The coefficient of static friction is typically larger than kinetic friction.

Question 17

A block sits on a rough horizontal surface. When a horizontal force of 20 N is applied, the block remains stationary. When the force is increased to 30 N, the block still doesn't move. At 35 N, the block just begins to slide. What is the magnitude of the friction force when the 20 N force is applied?

  1. 0 N
  2. 15 N
  3. 20 N (correct answer)
  4. 30 N
  5. 35 N
Explanation: When you encounter static friction problems, remember that static friction adjusts to match the applied force up to its maximum limit. The block remains in equilibrium until the applied force exceeds the maximum static friction. Let's analyze what happens at each force level. When 20 N is applied, the block stays stationary, meaning the net force is zero. Since static friction always opposes the applied force to maintain equilibrium, the friction force must equal 20 N in the opposite direction. Similarly, when 30 N is applied and the block still doesn't move, static friction increases to 30 N to maintain balance. The key insight comes from the 35 N case: this is when the block "just begins to slide," indicating we've reached the maximum static friction force of 35 N. Beyond this point, kinetic friction takes over and the block moves. Looking at the answer choices: A) 0 N is incorrect because with no friction force, the 20 N applied force would cause acceleration. B) 15 N is wrong because this would result in a net force of 5 N, causing motion. D) 30 N represents the friction force when 30 N is applied, not 20 N. C) 20 N is correct because static friction exactly balances the applied force. Remember this pattern: static friction equals the applied force (up to its maximum) when an object remains stationary. Always check whether the object is moving or stationary first, then apply the appropriate friction relationship. Static friction is variable, while kinetic friction is typically constant.

Question 18

A car traveling at 25 m/s begins to brake on a level road. The car's antilock braking system prevents the wheels from locking, maintaining kinetic friction between the tires and road. If the coefficient of kinetic friction is 0.75 and the car has a mass of 1200 kg, what is the minimum stopping distance?

  1. 42 m, found using the work-energy theorem with kinetic friction doing negative work (correct answer)
  2. 52 m, calculated assuming static friction provides the maximum braking force available
  3. 38 m, determined by applying Newton's second law to find deceleration then using kinematics
  4. 46 m, obtained by considering both kinetic friction and air resistance in the force analysis
Explanation: The maximum braking force is kinetic friction: fk=μkmg=0.75×1200×9.8=8820f_k = \mu_k mg = 0.75 \times 1200 \times 9.8 = 8820 N. This gives maximum deceleration: a=fk/m=8820/1200=7.35a = f_k/m = 8820/1200 = 7.35 m/s². Using kinematics: v2=u2+2asv^2 = u^2 + 2as with v=0v = 0, u=25u = 25 m/s, a=−7.35a = -7.35 m/s²: 0=(25)2+2(−7.35)s0 = (25)^2 + 2(-7.35)s, so s=625/14.7=42.5s = 625/14.7 = 42.5 m ≈ 42 m. Alternatively using work-energy theorem: work done against friction = initial kinetic energy: μkmg⋅d=12mv2\mu_k mg \cdot d = \frac{1}{2}mv^2, giving d=v2/(2μkg)=(25)2/(2×0.75×9.8)=625/14.7=42.5d = v^2/(2\mu_k g) = (25)^2/(2 \times 0.75 \times 9.8) = 625/14.7 = 42.5 m. Choice B incorrectly uses static friction, choice C mentions correct method but wrong value, choice D incorrectly includes air resistance.

Question 19

A 2.5 kg box is placed on a conveyor belt moving at constant speed. Initially, there is relative motion between the box and belt. The coefficient of kinetic friction is 0.35 and the coefficient of static friction is 0.45. Once the box reaches the same speed as the belt, what is the maximum acceleration the belt can have before the box begins to slip again?

  1. 3.4 m/s², limited by the maximum static friction force that can accelerate the box with the belt
  2. 4.4 m/s², determined by the maximum static friction available when no relative motion exists (correct answer)
  3. 2.8 m/s², calculated using kinetic friction since the box was initially sliding on the belt
  4. 5.2 m/s², found by considering the difference between static and kinetic friction coefficients
Explanation: Once the box and belt move together at the same speed, there is no relative motion, so static friction applies. The maximum static friction force provides the maximum force available to accelerate the box with the belt: fs,max=μsmg=0.45×2.5×9.8=11.0f_{s,max} = \mu_s mg = 0.45 \times 2.5 \times 9.8 = 11.0 N. Maximum acceleration: amax=fs,max/m=11.0/2.5=4.4a_{max} = f_{s,max}/m = 11.0/2.5 = 4.4 m/s². If the belt accelerates faster than this, the required force would exceed maximum static friction, causing the box to slip relative to the belt. Choice A uses wrong coefficient value, choice C incorrectly uses kinetic friction (which doesn't apply when there's no relative motion), choice D incorrectly attempts to use the difference between coefficients.

Question 20

A hockey puck slides across ice with an initial velocity of 8.0 m/s. Due to friction, it comes to rest after traveling 32 m. If the same puck is given an initial velocity of 12 m/s on the same ice surface, how far will it travel before coming to rest?

  1. 48 m, since distance is directly proportional to initial velocity when friction is constant
  2. 54 m, because the relationship between distance and velocity depends on the kinetic energy
  3. 72 m, found by applying the work-energy theorem with constant kinetic friction force (correct answer)
  4. 96 m, calculated using the assumption that friction force varies with velocity squared
Explanation: First scenario: Using v2=u2+2asv^2 = u^2 + 2as with v=0v = 0, u=8.0u = 8.0 m/s, s=32s = 32 m: 0=(8.0)2+2a(32)0 = (8.0)^2 + 2a(32), so a=−64/64=−1.0a = -64/64 = -1.0 m/s². The friction force causes this deceleration: μkg=1.0\mu_k g = 1.0, so μk=1.0/9.8=0.102\mu_k = 1.0/9.8 = 0.102. For the second scenario with u=12u = 12 m/s and same a=−1.0a = -1.0 m/s²: 0=(12)2+2(−1.0)s0 = (12)^2 + 2(-1.0)s, giving s=144/2=72s = 144/2 = 72 m. Alternatively, using energy: Work done against friction = initial kinetic energy. W=μkmg⋅d=12mv2W = \mu_k mg \cdot d = \frac{1}{2}mv^2. Since μk\mu_k and mm are constant, d∝v2d \propto v^2. So d2/d1=(v2/v1)2=(12/8)2=2.25d_2/d_1 = (v_2/v_1)^2 = (12/8)^2 = 2.25. Therefore d2=32×2.25=72d_2 = 32 \times 2.25 = 72 m.