College Physics Quiz: Resistive Forces
20 questions · exam conditions
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Resistive ForcesQuestion 1 of 20

A block slides down a rough inclined plane at constant velocity. If the angle of the incline is increased slightly while keeping all other conditions the same, what will happen to the block's motion immediately after the angle change?

The block will continue at constant velocity because the component of weight parallel to the incline equals the friction force.
The block will accelerate down the incline because the component of weight parallel to the incline now exceeds the maximum static friction force.
The block will accelerate down the incline because the component of weight parallel to the incline now exceeds the kinetic friction force.
The block will decelerate because the normal force decreases, reducing the available friction force to maintain constant velocity.
The block will maintain constant velocity because the coefficient of kinetic friction adjusts to balance the increased gravitational component.
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College Physics Quiz

College Physics Quiz: Resistive Forces

Practice Resistive Forces 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 Resistive Forces, 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.

All questions

Question 1

A block slides down a rough inclined plane at constant velocity. If the angle of the incline is increased slightly while keeping all other conditions the same, what will happen to the block's motion immediately after the angle change?

  1. The block will continue at constant velocity because the component of weight parallel to the incline equals the friction force.
  2. The block will accelerate down the incline because the component of weight parallel to the incline now exceeds the maximum static friction force.
  3. The block will accelerate down the incline because the component of weight parallel to the incline now exceeds the kinetic friction force. (correct answer)
  4. The block will decelerate because the normal force decreases, reducing the available friction force to maintain constant velocity.
  5. The block will maintain constant velocity because the coefficient of kinetic friction adjusts to balance the increased gravitational component.
Explanation: When you encounter problems involving objects on inclined planes, focus on analyzing the forces and how changes in geometry affect the force balance. Initially, the block slides at constant velocity, meaning the net force is zero. The component of weight parallel to the incline (mgsin⁡θmg\sin\theta) exactly equals the kinetic friction force (μkN=μkmgcos⁡θ\mu_k N = \mu_k mg\cos\theta). This establishes the coefficient of kinetic friction: μk=tan⁡θ\mu_k = \tan\theta. When the angle increases slightly, the parallel component of weight increases to mgsin⁡(θ+Δθ)mg\sin(\theta + \Delta\theta), while the normal force decreases to mgcos⁡(θ+Δθ)mg\cos(\theta + \Delta\theta). Since the block was already sliding, kinetic friction still applies, but now the kinetic friction force becomes μkmgcos⁡(θ+Δθ)\mu_k mg\cos(\theta + \Delta\theta). Because sin⁡(θ+Δθ)>sin⁡θ\sin(\theta + \Delta\theta) > \sin\theta and cos⁡(θ+Δθ)<cos⁡θ\cos(\theta + \Delta\theta) < \cos\theta, the parallel weight component increases while the friction force decreases. This creates a net force down the incline, causing acceleration. Choice A incorrectly assumes equilibrium continues after the angle change. Choice B mentions "maximum static friction," but static friction doesn't apply since the block is already in motion—this is a kinetic friction situation. Choice D correctly notes that normal force decreases, but incorrectly suggests this causes deceleration when it actually contributes to acceleration by reducing the opposing friction force. Remember: once an object is sliding on an inclined plane, any increase in angle will cause acceleration because it simultaneously increases the driving force and decreases the opposing friction force.

Question 2

A car travels at constant speed around a horizontal circular track. The centripetal force is provided entirely by friction between the tires and road. If the car's speed doubles while the radius remains constant, by what factor does the required coefficient of static friction change?

  1. The coefficient increases by a factor of 2 because the centripetal acceleration doubles with speed.
  2. The coefficient increases by a factor of 4 because the centripetal force depends on the square of the velocity. (correct answer)
  3. The coefficient increases by a factor of 2\sqrt{2} because the relationship between friction and velocity is nonlinear.
  4. The coefficient remains the same because it depends only on the surfaces in contact, not on the motion.
  5. The coefficient decreases by a factor of 2 because the normal force increases with the centripetal acceleration.
Explanation: When analyzing circular motion problems, you need to connect centripetal force requirements with the forces actually available to provide that acceleration. For circular motion, the centripetal force is Fc=mv2rF_c = \frac{mv^2}{r}. Since friction provides this force, we have f=μsmg=mv2rf = \mu_s mg = \frac{mv^2}{r}. Solving for the coefficient of static friction: μs=v2rg\mu_s = \frac{v^2}{rg}. Notice that the coefficient of friction depends on v2v^2 - the square of the velocity. When the car's speed doubles (vv becomes 2v2v), the required coefficient becomes μs=(2v)2rg=4v2rg\mu_s = \frac{(2v)^2}{rg} = \frac{4v^2}{rg}. This is exactly 4 times the original coefficient, confirming that answer B is correct. Let's examine why the other options fail: Choice A incorrectly states that centripetal acceleration doubles with speed, but acceleration actually increases as v2v^2, not linearly with vv. Choice C suggests a square root relationship that doesn't exist in the physics - the relationship is clearly quadratic, not involving 2\sqrt{2}. Choice D reflects a common misconception that the coefficient of friction is purely a material property, but here we're finding the required coefficient to maintain circular motion, which absolutely depends on the motion parameters. Remember: In circular motion problems, always check whether force requirements scale linearly or quadratically with speed. Centripetal force (and anything providing it) scales as v2v^2, making quadratic relationships very common in these scenarios.

Question 3

A skydiver reaches terminal velocity during free fall. At this point, which statement correctly describes the forces acting on the skydiver?

  1. The gravitational force equals the air resistance force, and both forces are directed vertically downward toward Earth.
  2. The gravitational force exceeds the air resistance force by exactly the amount needed to maintain constant acceleration.
  3. The gravitational force equals the air resistance force in magnitude, with gravity acting downward and air resistance acting upward. (correct answer)
  4. The gravitational force is partially canceled by air resistance, leaving a small net downward force to maintain motion.
  5. The air resistance force slightly exceeds the gravitational force to prevent further acceleration and maintain equilibrium.
Explanation: When analyzing motion problems involving terminal velocity, you need to apply Newton's first law: an object at constant velocity has zero net force acting on it. At terminal velocity, the skydiver moves downward at constant speed, meaning acceleration is zero. For this to happen, all forces must be perfectly balanced. Two forces act on the skydiver: gravitational force (weight) pulling downward, and air resistance (drag) pushing upward against the motion. These forces must be equal in magnitude but opposite in direction, creating zero net force. Answer C correctly states this balance: gravitational force equals air resistance force in magnitude, with gravity acting downward and air resistance acting upward. The forces cancel completely. Answer A incorrectly claims both forces act downward. Air resistance always opposes motion direction, so it acts upward during downward fall. If both forces pointed downward, the skydiver would accelerate continuously. Answer B suggests gravitational force exceeds air resistance to maintain "constant acceleration." This contradicts terminal velocity's definition—constant velocity means zero acceleration, not constant acceleration. Answer D states air resistance only "partially cancels" gravity, leaving a net downward force. Any net force would cause acceleration, preventing terminal velocity from being achieved. Study tip: Remember that terminal velocity means zero acceleration, which requires zero net force. Whenever you see "terminal velocity" or "constant velocity," immediately think "balanced forces." This principle applies to any object moving at constant speed, whether falling, sliding, or floating.

Question 4

A hockey puck sliding on ice experiences air resistance proportional to its velocity: Fair=−bvF_{air} = -bv, where bb is a positive constant. As the puck slows down, which statement correctly describes how the magnitude of air resistance changes?

  1. Air resistance increases as the puck slows because the relative motion between puck and air becomes more significant.
  2. Air resistance decreases linearly with velocity, so the deceleration of the puck decreases over time as well. (correct answer)
  3. Air resistance remains constant because it depends on the surface area and air density, not velocity.
  4. Air resistance decreases exponentially because the drag force follows an inverse relationship with time.
  5. Air resistance approaches zero as the velocity approaches zero, but the deceleration remains constant throughout the motion.
Explanation: This problem tests your understanding of velocity-dependent drag forces and how they affect motion over time. When you encounter questions about air resistance or drag, focus on the mathematical relationship given and trace through its physical consequences. The key insight is in the given equation: Fair=−bvF_{air} = -bv. Since air resistance is directly proportional to velocity, as the puck slows down (velocity decreases), the magnitude of air resistance must also decrease proportionally. This creates a feedback effect: as the puck gets slower, air resistance gets weaker, which means the deceleration becomes less severe over time. Using Newton's second law, ma=−bvma = -bv, so the acceleration is a=−bvma = -\frac{bv}{m}. Since acceleration represents the rate of velocity change, smaller velocities produce smaller decelerations. This confirms that answer B is correct—both air resistance and deceleration decrease as the puck slows. Answer A incorrectly suggests air resistance increases, contradicting the linear relationship F=−bvF = -bv. Answer C wrongly claims air resistance is constant, ignoring the velocity dependence explicitly given in the problem. Answer D mentions exponential decrease, which confuses the time-dependence of velocity (which does decay exponentially in this scenario) with the force-velocity relationship, which is linear. Remember: when drag force is proportional to velocity, always trace through the chain of effects. Slower speed → less drag → less deceleration → gradually approaching (but never quite reaching) zero velocity. This creates the characteristic exponential decay in velocity you'll see in many real-world situations.

Question 5

A car brakes to a stop on a level road. The maximum deceleration depends on the coefficient of static friction between the tires and road. If the car's mass is doubled but all other conditions remain the same, how does this affect the maximum possible deceleration?

  1. The maximum deceleration is halved because the increased mass requires more force to achieve the same acceleration.
  2. The maximum deceleration doubles because the increased weight provides more normal force for friction.
  3. The maximum deceleration remains the same because both the available friction force and the mass increase proportionally. (correct answer)
  4. The maximum deceleration increases by a factor of 2\sqrt{2} due to the nonlinear relationship between mass and friction.
  5. The maximum deceleration becomes unpredictable because it depends on the specific braking technique used by the driver.
Explanation: When analyzing braking problems, you need to understand the relationship between friction, normal force, and Newton's second law. The maximum deceleration occurs when the friction force reaches its maximum value, which happens at the threshold of slipping. The maximum friction force is fmax=μsNf_{max} = \mu_s N, where μs\mu_s is the coefficient of static friction and NN is the normal force. On a level road, the normal force equals the weight: N=mgN = mg. So the maximum friction force becomes fmax=μsmgf_{max} = \mu_s mg. Using Newton's second law, F=maF = ma, the maximum deceleration is: amax=fmaxm=μsmgm=μsga_{max} = \frac{f_{max}}{m} = \frac{\mu_s mg}{m} = \mu_s g. Notice that mass cancels out completely! The maximum deceleration depends only on the coefficient of friction and gravitational acceleration. Choice A incorrectly assumes that doubling mass reduces deceleration without recognizing that the available friction force also doubles. This misses the proportional relationship. Choice B makes the opposite error, thinking that more normal force somehow increases deceleration beyond what the mass increase would offset. While more normal force does create more friction, this extra friction is exactly what's needed to decelerate the larger mass. Choice D introduces an imaginary nonlinear relationship that doesn't exist in the friction equations. Remember this key insight: maximum deceleration due to friction is independent of mass because both the friction force and the inertia increase proportionally. This same principle applies to any object sliding down an inclined plane or experiencing friction-limited motion.

Question 6

A block slides down a frictionless incline and then onto a horizontal surface with kinetic friction. The transition between surfaces is smooth. At the moment the block reaches the horizontal surface, which forces act on the block?

  1. Weight acting downward, normal force from the horizontal surface acting upward, and kinetic friction acting opposite to the velocity direction. (correct answer)
  2. Weight acting downward, normal forces from both the incline and horizontal surface, and kinetic friction acting horizontally backward.
  3. Weight acting downward, normal force acting perpendicular to the incline, and kinetic friction acting up the incline to oppose motion.
  4. Weight acting downward, normal force acting upward, kinetic friction acting horizontally backward, and a residual force component from the incline.
  5. Only weight and kinetic friction, since the normal force transitions gradually between the inclined and horizontal orientations.
Explanation: When analyzing forces on an object, you must identify which surfaces are actually in contact with the object at that specific moment. Forces only exist when there's direct physical interaction. At the moment the block reaches the horizontal surface, it's transitioning from the incline to the flat ground. The block is now in contact with the horizontal surface, so three forces act on it: its weight (gravitational force) pulls downward, the normal force from the horizontal surface pushes upward perpendicular to that surface, and kinetic friction opposes its motion horizontally backward. The normal force adjusts to balance the weight vertically, while friction opposes the horizontal velocity component. Option A correctly identifies these three forces with their proper directions. Option B incorrectly suggests normal forces from both surfaces act simultaneously - but the block can't experience forces from the incline when it's no longer touching the incline. Option C mistakenly places the normal force perpendicular to the incline and friction up the incline, which would only be true if the block were still on the inclined surface. Option D introduces a "residual force component from the incline," which violates the fundamental principle that forces require contact - once the block leaves the incline, no forces can originate from it. Remember: forces only exist through direct contact or field interactions (like gravity). When an object transitions between surfaces, carefully identify which surfaces are actually touching the object at that specific instant, not where it was or will be.

Question 7

Two identical blocks are placed on inclined planes with the same angle but different coefficients of kinetic friction. Block A slides down at constant velocity while Block B accelerates down the incline. If both blocks have the same mass mm and the incline angle is θ\theta, what can be concluded about the friction forces?

  1. The friction force on Block A is greater than on Block B because Block A maintains constant velocity.
  2. The friction force on Block A equals mgsin⁡θmg\sin\theta while the friction force on Block B is less than mgsin⁡θmg\sin\theta. (correct answer)
  3. Both blocks experience the same friction force because they have the same mass and normal force.
  4. The friction force on Block B is greater because acceleration requires additional force beyond static equilibrium.
  5. The friction forces cannot be compared without knowing the specific values of the coefficients of kinetic friction.
Explanation: When analyzing forces on inclined planes, you need to consider whether objects are in equilibrium (constant velocity) or experiencing net acceleration, as this determines the relationship between gravitational and friction forces. For Block A moving at constant velocity, Newton's first law tells us the net force is zero. The component of gravitational force down the incline is mgsin⁡θmg\sin\theta, so the friction force must exactly balance this: fA=mgsin⁡θf_A = mg\sin\theta. For Block B accelerating down the incline, there's a net downward force, meaning gravity exceeds friction: mgsin⁡θ>fBmg\sin\theta > f_B, so fB<mgsin⁡θf_B < mg\sin\theta. Choice A incorrectly suggests Block A has greater friction force. While Block A's friction equals mgsin⁡θmg\sin\theta, Block B's friction is actually smaller, allowing acceleration to occur. Choice C makes the common error of assuming equal normal forces automatically mean equal friction forces. While both blocks do have the same normal force (mgcos⁡θmg\cos\theta), they have different coefficients of kinetic friction, so f=μkNf = \mu_k N gives different friction forces. Choice D misunderstands the physics entirely—Block B accelerates precisely because its friction force is weaker, not stronger, than the gravitational component. Choice B correctly identifies that Block A's friction equals mgsin⁡θmg\sin\theta (equilibrium condition) while Block B's friction is less than mgsin⁡θmg\sin\theta (allowing net acceleration). Study tip: In inclined plane problems, constant velocity always means forces balance exactly, while acceleration means one force dominates. Use this to quickly determine force relationships before doing detailed calculations.

Question 8

A wooden block is pulled along a horizontal surface by a string. The tension in the string is 25 N, the block moves at constant velocity, and the coefficient of kinetic friction between block and surface is 0.30. What is the weight of the block?

  1. 75 N, because the friction force equals the coefficient times the weight, and friction balances the tension.
  2. 83 N, because the normal force must account for both the weight and the horizontal pulling force effects.
  3. 25 N, because the weight equals the tension required to overcome friction at constant velocity.
  4. 125 N, because the tension represents only a fraction of the total weight needed to overcome static friction.
  5. Cannot be determined without knowing the angle at which the string pulls the block. (correct answer)
Explanation: When you encounter problems involving objects moving at constant velocity with friction, remember that all forces must be balanced. This is a direct application of Newton's first law. Since the block moves at constant velocity, the net force is zero. This means the tension force (25 N) pulling the block forward exactly equals the kinetic friction force opposing the motion. The kinetic friction force is given by fk=μkNf_k = \mu_k N, where μk=0.30\mu_k = 0.30 and NN is the normal force. On a horizontal surface with no vertical forces other than weight and normal force, the normal force equals the weight of the block. Setting up the force balance: tension = friction force, so 25=0.30×W25 = 0.30 \times W, where WW is the weight. Solving: W=250.30=83.3W = \frac{25}{0.30} = 83.3 N, which rounds to 83 N. Looking at the wrong answers: Choice A incorrectly assumes the weight is 75 N without proper calculation. Choice C confuses tension with weight, missing that friction depends on the normal force. Choice D incorrectly brings up static friction and suggests an unrealistic relationship between tension and total weight. Choice B correctly identifies that we need 83 N, though its reasoning about "horizontal pulling force effects" is poorly worded - the key insight is simply that normal force equals weight on a horizontal surface. Strategy tip: In constant velocity problems, immediately set up force balance equations. Horizontal forces balance separately from vertical forces, and kinetic friction always equals μk\mu_k times the normal force.

Question 9

A race car travels around a banked circular track at constant speed. The banking angle is designed so that no friction is needed to provide the centripetal force at a specific speed. If the car travels faster than this design speed, what provides the additional centripetal force needed?

  1. The normal force from the track increases automatically to provide the additional centripetal force required.
  2. Static friction acts radially inward (toward the center of the track) to supplement the centripetal force from banking. (correct answer)
  3. The car's weight effectively increases due to the higher speed, providing additional downward force for banking.
  4. Static friction acts radially outward (away from the center) to prevent the car from sliding up the banking.
  5. The banking becomes more effective at higher speeds due to increased normal force from centripetal acceleration.
Explanation: When analyzing banked curves in circular motion, you need to understand how forces combine to provide the required centripetal force. At the design speed, the horizontal component of the normal force alone provides exactly the right centripetal force, with no friction needed. When the car exceeds this design speed, it needs more centripetal force than banking alone can provide. The car would naturally tend to slide up and outward on the banked surface due to insufficient inward force. To prevent this upward sliding, static friction must act down the incline, which has a component pointing radially inward toward the track's center. This inward friction component supplements the centripetal force from banking. Let's examine why the other options fail: (A) is incorrect because the normal force magnitude depends on the car's weight and the banking angle, not directly on speed. The normal force doesn't automatically adjust to provide whatever centripetal force is needed. (C) misunderstands physics entirely—a car's weight (gravitational force) doesn't change with horizontal speed. Weight depends only on mass and gravitational acceleration. (D) describes friction acting outward, which would actually reduce the available centripetal force rather than increase it. Remember this key insight: on banked curves, friction always acts to prevent the car from sliding in whatever direction it would naturally go. Above design speed, cars tend to slide outward and upward, so friction acts inward and downward, contributing additional centripetal force.

Question 10

A block slides up a rough inclined plane with initial velocity v0v_0. Due to both gravity and friction, it comes to rest and then slides back down. Comparing the upward and downward portions of motion, which statement is correct?

  1. The magnitude of acceleration is greater during the upward motion because both gravity and friction oppose the motion. (correct answer)
  2. The magnitude of acceleration is greater during the downward motion because gravity assists while friction still opposes motion.
  3. The magnitude of acceleration is the same for both portions because the same forces act on the block throughout.
  4. The acceleration during upward motion is g(sin⁡θ+μkcos⁡θ)g(\sin\theta + \mu_k\cos\theta) while during downward motion it is g(sin⁡θ−μkcos⁡θ)g(\sin\theta - \mu_k\cos\theta).
  5. The accelerations cannot be compared without knowing the specific angle of the incline and coefficient of friction.
Explanation: When analyzing motion on inclined planes with friction, you need to carefully consider how the direction of motion affects the friction force, which always opposes the actual direction of movement. During upward motion, the block moves against gravity (which pulls down the incline) and against kinetic friction (which opposes the upward motion by acting down the incline). Both forces point in the same direction down the incline, so the acceleration magnitude is aup=g(sin⁡θ+μkcos⁡θ)a_{up} = g(\sin\theta + \mu_k\cos\theta). During downward motion, gravity still pulls down the incline, but now friction opposes the downward motion by acting up the incline. These forces oppose each other, giving adown=g(sin⁡θ−μkcos⁡θ)a_{down} = g(\sin\theta - \mu_k\cos\theta). Since μkcos⁡θ\mu_k\cos\theta is always positive, g(sin⁡θ+μkcos⁡θ)>g(sin⁡θ−μkcos⁡θ)g(\sin\theta + \mu_k\cos\theta) > g(\sin\theta - \mu_k\cos\theta), making the upward acceleration magnitude greater. Looking at the wrong answers: Option B incorrectly suggests downward acceleration is greater, missing that friction's direction reverses. Option C wrongly claims the accelerations are equal, ignoring that friction changes direction while maintaining the same magnitude. Option D correctly states the acceleration expressions but doesn't compare their magnitudes to answer which is greater. Therefore, A is correct: the acceleration magnitude is greater during upward motion because both gravity and friction oppose the motion. Study tip: Remember that kinetic friction always opposes the direction of actual motion, not the direction of other forces. When motion direction changes, friction direction changes too.

Question 11

A parachutist jumps from an airplane and initially accelerates downward. As speed increases, air resistance increases until terminal velocity is reached. During the transition from initial acceleration to terminal velocity, which best describes the motion?

  1. The acceleration decreases linearly with time as air resistance increases proportionally to the falling distance.
  2. The acceleration decreases continuously as air resistance increases, approaching zero acceleration as terminal velocity is approached. (correct answer)
  3. The acceleration remains constant until terminal velocity is suddenly reached when air resistance equals the weight.
  4. The acceleration first increases due to gravity, then decreases as air resistance builds up to balance the weight.
  5. The acceleration oscillates around zero as the parachutist alternately speeds up and slows down approaching terminal velocity.
Explanation: When analyzing falling objects with air resistance, you need to understand how forces change during the fall and how this affects acceleration. At the moment of jumping, the parachutist experiences maximum acceleration downward because air resistance is initially zero (or very small) while gravitational force remains constant. As speed increases, air resistance grows stronger, typically proportional to velocity squared. This creates an upward force that opposes gravity. The correct answer is B because Newton's second law tells us that net force determines acceleration: Fnet=mg−Fair=maF_{net} = mg - F_{air} = ma. As air resistance increases, the net downward force decreases, so acceleration must decrease continuously. Eventually, when air resistance equals weight (Fair=mgF_{air} = mg), net force becomes zero, meaning acceleration reaches zero and terminal velocity is achieved. Answer A is incorrect because acceleration doesn't decrease linearly with time, and air resistance depends on velocity, not falling distance. The relationship is more complex due to the velocity-squared dependence of air resistance. Answer C is wrong because acceleration changes gradually, not suddenly. Terminal velocity is approached asymptotically - the parachutist gets closer and closer to it but never suddenly "reaches" it. Answer D is incorrect because acceleration doesn't first increase. Gravity provides constant force throughout the fall, so the maximum acceleration occurs at the very beginning when air resistance is minimal. Remember: when analyzing motion with air resistance, focus on how the opposing forces change with velocity and how this affects the net force and resulting acceleration.

Question 12

A heavy crate is pushed across a warehouse floor. Initially, a large force is required to get it moving, but once moving, a smaller force maintains constant velocity. This observation illustrates the difference between which two types of friction?

  1. Kinetic friction and rolling friction, where rolling friction is always less than kinetic friction for the same surfaces.
  2. Static friction and kinetic friction, where the maximum static friction typically exceeds kinetic friction for the same surfaces. (correct answer)
  3. Internal friction and external friction, where internal friction within the crate decreases once motion begins.
  4. Coulomb friction and viscous friction, where viscous friction dominates at higher speeds while Coulomb friction dominates at rest.
  5. Dry friction and lubricated friction, where the transition occurs when motion generates heat that lubricates the surfaces.
Explanation: When analyzing forces needed to move objects, you're examining different types of friction that act under different conditions. The key insight is understanding how friction changes between stationary and moving states. The scenario describes a classic demonstration of static versus kinetic friction. Static friction acts when surfaces are not sliding relative to each other - it's what keeps the crate at rest initially. This force can vary from zero up to a maximum value (μsN\mu_s N, where μs\mu_s is the coefficient of static friction and NN is the normal force). Once you overcome maximum static friction, the crate begins moving and kinetic friction takes over. Kinetic friction (μkN\mu_k N) typically has a lower coefficient than static friction, explaining why less force is needed to maintain motion than to initiate it. This makes answer B correct. Answer A confuses kinetic friction with rolling friction, which involves wheels or cylinders rolling rather than sliding. Answer C incorrectly suggests internal friction within the crate itself, but the friction occurs between the crate and floor surfaces. Answer D references Coulomb friction (dry friction between solid surfaces) and viscous friction (fluid resistance), but viscous effects aren't relevant for a crate on a warehouse floor, and the speed-dependent behavior described doesn't match the scenario. Remember this key pattern: if a physics problem describes needing more force to start motion than to maintain it, you're dealing with the static-to-kinetic friction transition. The coefficients almost always follow μs>μk\mu_s > \mu_k.

Question 13

A block rests on a rough inclined plane. The angle of the incline is gradually increased from zero. At what point will the block begin to slide, and what force relationship determines this critical angle?

  1. The block slides when the component of weight parallel to the incline equals the component perpendicular to the incline.
  2. The block slides when the component of weight parallel to the incline exceeds the maximum static friction force available. (correct answer)
  3. The block slides when the normal force becomes less than the component of weight parallel to the incline.
  4. The block slides when the coefficient of static friction equals the sine of the incline angle.
  5. The block slides when the total weight of the block exceeds the maximum friction force that can be generated.
Explanation: When analyzing objects on inclined planes, you need to understand the force balance that determines when sliding begins. This involves comparing the driving force (gravity component) with the resisting force (friction). As the incline angle increases, the component of the block's weight parallel to the incline, mgsin⁡θmg\sin\theta, also increases. This parallel component is what tries to pull the block down the slope. Meanwhile, static friction acts up the incline to resist this motion. The maximum static friction available is fs,max=μsN=μsmgcos⁡θf_{s,max} = \mu_s N = \mu_s mg\cos\theta, where μs\mu_s is the coefficient of static friction and NN is the normal force. The block begins sliding when the parallel weight component exceeds the maximum static friction force: mgsin⁡θ>μsmgcos⁡θmg\sin\theta > \mu_s mg\cos\theta. This confirms that answer B is correct – sliding occurs when the driving force overcomes the maximum available friction. Answer A incorrectly suggests sliding happens when parallel and perpendicular weight components are equal (sin⁡θ=cos⁡θ\sin\theta = \cos\theta), which would occur at 45° regardless of surface roughness – clearly wrong since a rougher surface should resist sliding better. Answer C confuses force directions by comparing normal force with parallel weight component – these forces are perpendicular to each other and this comparison doesn't determine sliding. Answer D states μs=sin⁡θ\mu_s = \sin\theta, but the actual critical condition is μs=tan⁡θ\mu_s = \tan\theta (derived from mgsin⁡θ=μsmgcos⁡θmg\sin\theta = \mu_s mg\cos\theta). Remember: sliding begins when the driving force exceeds maximum static friction. Always identify what's trying to cause motion versus what's resisting it.

Question 14

A box is pulled across a horizontal surface by a rope that makes a 30° angle above the horizontal. The box moves at constant velocity. If the pulling force is 50 N, the coefficient of kinetic friction is 0.40, and the box weighs 80 N, what is the normal force between the box and the surface?

  1. 55 N because the normal force equals the weight minus the upward component of the pulling force. (correct answer)
  2. 80 N because the normal force always equals the weight of the object on a horizontal surface.
  3. 105 N because the normal force equals the weight plus the component of the pulling force.
  4. 65 N because the normal force must account for the vertical equilibrium including friction effects.
  5. 45 N because the kinetic friction reduces the effective normal force by the horizontal component of the pull.
Explanation: When analyzing forces on an object being pulled at an angle, you need to consider how the angled force affects the normal force. The key insight is that any upward component of the pulling force reduces the contact force between the object and surface. Since the box moves at constant velocity, all forces are in equilibrium. Let's examine the vertical forces: the weight (80 N downward), the normal force (upward), and the vertical component of the pulling force. The pulling force of 50 N at 30° has a vertical component of 50sin⁡(30°)=50×0.5=2550 \sin(30°) = 50 × 0.5 = 25 N upward. For vertical equilibrium: Normal force + 25 N = 80 N, so the normal force = 80 N - 25 N = 55 N. Looking at the wrong answers: Choice B incorrectly assumes the normal force always equals the weight on horizontal surfaces. This is only true when no other vertical forces act on the object. Choice C adds the force component instead of subtracting it, which would increase the normal force rather than decrease it. Choice D mentions friction effects on the normal force, but friction acts horizontally and doesn't directly affect the vertical force balance. The correct answer is A because it properly recognizes that the upward component of the angled pulling force reduces the normal force below the object's weight. Study tip: When forces act at angles, always break them into components first. The normal force equals the weight only when no other vertical forces are present - any upward force reduces it, any downward force increases it.

Question 15

A box sits on the floor of an elevator. As the elevator starts moving upward from rest, it accelerates upward at 2.0 m/s22.0 \text{ m/s}^2. If the coefficient of static friction between the box and elevator floor is 0.40, what is the maximum horizontal force that can be applied to the box without causing it to slip?

  1. The maximum force is 0.40mg0.40mg because the friction coefficient determines the maximum horizontal force regardless of vertical acceleration.
  2. The maximum force is 0.40m(g+2.0)0.40m(g + 2.0) because the upward acceleration effectively increases the normal force between box and floor. (correct answer)
  3. The maximum force is 0.40m(g−2.0)0.40m(g - 2.0) because the upward acceleration reduces the contact force between box and floor.
  4. The maximum force cannot be determined without knowing whether the applied horizontal force accelerates the elevator or just the box.
  5. The maximum force is 0.40mg2+(2.0)20.40m\sqrt{g^2 + (2.0)^2} because the effective acceleration combines vertical and horizontal components.
Explanation: When you encounter problems involving friction in accelerating reference frames, you need to carefully analyze how the acceleration affects the normal force, which in turn determines the maximum friction force available. Since the elevator accelerates upward at 2.0 m/s22.0 \text{ m/s}^2, the box experiences an effective acceleration in the same direction. To find the normal force, apply Newton's second law in the vertical direction. The forces on the box are its weight mgmg (downward) and the normal force NN from the floor (upward). Since the box accelerates upward with the elevator: N−mg=maN - mg = ma, where a=2.0 m/s2a = 2.0 \text{ m/s}^2. Therefore, N=m(g+a)=m(g+2.0)N = m(g + a) = m(g + 2.0). The maximum static friction force equals μsN=0.40×m(g+2.0)=0.40m(g+2.0)\mu_s N = 0.40 \times m(g + 2.0) = 0.40m(g + 2.0). This is the maximum horizontal force that can be applied before slipping occurs, confirming answer B. Answer A incorrectly assumes the normal force remains mgmg, ignoring the effect of upward acceleration. This would only be true if the elevator weren't accelerating. Answer C uses the wrong sign for acceleration, suggesting the upward acceleration somehow reduces the normal force—this would only happen if the elevator were accelerating downward. Answer D introduces an irrelevant distinction about what the force accelerates; the question simply asks for the maximum horizontal force before slipping. Remember: in accelerating reference frames, always recalculate the normal force using Newton's second law before determining friction limits. The normal force isn't always equal to the object's weight.

Question 16

A box is pushed across a rough horizontal surface at constant velocity by a horizontal force of 40 N. If the same box is instead pulled by a rope making a 45° angle above the horizontal, what pulling force is required to maintain the same constant velocity?

  1. 40 N, because the same net horizontal force is required regardless of the force direction.
  2. Less than 40 N, because the upward component of the pull reduces the normal force and thus the friction. (correct answer)
  3. More than 40 N, because only the horizontal component of the pull contributes to overcoming friction.
  4. 40240\sqrt{2} N, because the force must be increased by the secant of the angle to maintain the same horizontal component.
  5. 402\frac{40}{\sqrt{2}} N, because the diagonal pull is more efficient than the horizontal push due to the reduced normal force.
Explanation: When analyzing forces on moving objects, always start by identifying what forces are acting and how changing the force direction affects the force balance. In the first scenario, the 40 N horizontal push exactly balances the friction force, so the box moves at constant velocity. When you switch to pulling at a 45° angle, two key changes occur that affect the required force. The pulling force now has two components: horizontal (Fcos⁡45°F\cos 45°) and vertical (Fsin⁡45°F\sin 45°). The horizontal component must still overcome friction, but here's the crucial insight: the upward component reduces the normal force between the box and surface. Since friction equals the coefficient of friction times the normal force (f=μNf = \mu N), reducing the normal force reduces the friction force that needs to be overcome. Let's say the required pulling force is FF. Then Fcos⁡45°=μ(mg−Fsin⁡45°)F\cos 45° = \mu(mg - F\sin 45°). Since cos⁡45°=sin⁡45°=12\cos 45° = \sin 45° = \frac{1}{\sqrt{2}}, we get F2=μ(mg−F2)\frac{F}{\sqrt{2}} = \mu(mg - \frac{F}{\sqrt{2}}). Solving this shows F<40F < 40 N. Answer A incorrectly assumes only horizontal forces matter, ignoring how the vertical component affects normal force. Answer C makes the opposite error, recognizing the horizontal component limitation but missing that reduced friction makes the job easier. Answer D applies the secant incorrectly—this would give the force needed for the same horizontal component, but ignores that less horizontal force is actually needed due to reduced friction. Remember: when forces change direction, always consider how all components affect the entire force system, not just the obvious horizontal effects.

Question 17

A ball is thrown vertically upward in air. Considering air resistance that opposes motion, which statement correctly describes the forces during the upward portion of the flight?

  1. Both gravity and air resistance act downward, causing the ball to decelerate more rapidly than in vacuum. (correct answer)
  2. Gravity acts downward while air resistance acts upward, partially reducing the net deceleration compared to vacuum.
  3. Gravity and air resistance alternate directions depending on the ball's instantaneous acceleration.
  4. Air resistance acts tangentially to the trajectory while gravity acts radially, creating complex motion patterns.
  5. The net effect of air resistance is negligible during upward motion because the ball moves against natural air currents.
Explanation: When analyzing projectile motion with air resistance, you need to carefully consider the direction of each force relative to the object's velocity, not just its position. During the upward flight, the ball moves upward while two forces act on it. Gravity always pulls downward with force mgmg. Air resistance always opposes the direction of motion - since the ball moves upward, air resistance acts downward as well. These forces combine to create a net downward force greater than gravity alone, causing the ball to decelerate more rapidly than it would in a vacuum. Option A correctly identifies this scenario. Both forces act downward, creating greater deceleration than in vacuum. Option B makes a critical error about air resistance direction. It assumes air resistance acts upward, but air resistance always opposes motion. Since the ball moves upward, air resistance must act downward, not upward. Option C incorrectly suggests forces change direction based on acceleration. Force directions depend on fundamental interactions (gravity always down) and motion direction (air resistance opposite to velocity), not on the resulting acceleration. Option D misapplies concepts from circular motion. In projectile motion, both gravity and air resistance act along the vertical direction - there's no radial or tangential component since the trajectory, while curved, doesn't involve rotation about a fixed center. Remember: Air resistance always opposes the direction of motion, not the direction of other forces. When analyzing motion with air resistance, first identify the velocity direction, then determine that air resistance acts opposite to it.

Question 18

A child slides down a playground slide that curves from steep to shallow. The coefficient of kinetic friction is constant throughout. As the child moves from the steep section to the shallow section while maintaining contact with the slide, how does the friction force change?

  1. Friction force increases because the normal force increases as the slide becomes less steep. (correct answer)
  2. Friction force decreases because less friction is needed to oppose motion on the shallow section.
  3. Friction force remains constant because the coefficient of kinetic friction and the child's weight are unchanged.
  4. Friction force first decreases then increases due to the changing velocity as the child transitions between sections.
  5. Friction force becomes zero momentarily during the transition because the normal force changes direction.
Explanation: When analyzing friction on inclined surfaces, you need to understand how the normal force changes with the angle of inclination. The friction force depends on both the coefficient of kinetic friction and the normal force: fk=μkNf_k = \mu_k N. As the slide transitions from steep to shallow, the angle with the horizontal decreases. The normal force equals the component of the child's weight perpendicular to the surface: N=mgcos⁡θN = mg\cos\theta. When the angle θ\theta becomes smaller (less steep), cos⁡θ\cos\theta increases, making the normal force larger. Since the coefficient of kinetic friction remains constant, a larger normal force means a larger friction force. Looking at the wrong answers: Option B incorrectly assumes friction adjusts based on what's "needed" - but kinetic friction depends only on the normal force and coefficient, not on motion requirements. Option C misses that while the coefficient and weight are constant, the normal force changes because it's only the perpendicular component of weight. Option D suggests friction varies with velocity changes, but kinetic friction is independent of speed - it only depends on the normal force. The key insight is that option A correctly identifies that the normal force increases as the surface becomes less steep, directly increasing the friction force through the relationship fk=μkNf_k = \mu_k N. Study tip: For inclined plane problems, always break weight into components parallel and perpendicular to the surface. Remember that kinetic friction depends only on the normal force and coefficient - it's not about what force is "needed" or the object's speed.

Question 19

A 1500 kg car travels at constant speed around a horizontal circular track of radius 200 m. The coefficient of static friction between the tires and track is 0.75, and the coefficient of kinetic friction is 0.60. If the car gradually increases its speed, at what speed will the tires first begin to slip?

  1. 34.3 m/s34.3 \text{ m/s}
  2. 38.4 m/s38.4 \text{ m/s} (correct answer)
  3. 42.0 m/s42.0 \text{ m/s}
  4. 29.4 m/s29.4 \text{ m/s}
Explanation: The tires will begin to slip when the required centripetal force exceeds the maximum static friction force. The centripetal force required is Fc=mv2/rF_c = mv^2/r, and the maximum static friction force is fs,max=μsmgf_{s,max} = \mu_s mg. At the point of slipping: mv2/r=μsmgmv^2/r = \mu_s mg. Solving for vv: v2=μsgr=0.75×9.8×200=1470v^2 = \mu_s gr = 0.75 \times 9.8 \times 200 = 1470, so v=1470=38.34 m/sv = \sqrt{1470} = 38.34 \text{ m/s}. This rounds to 38.4 m/s. Choice A (34.3 m/s) would result from using kinetic friction: v=0.60×9.8×200=1176=34.3 m/sv = \sqrt{0.60 \times 9.8 \times 200} = \sqrt{1176} = 34.3 \text{ m/s}, but this is incorrect since slipping begins when static friction is overcome. Choice C (42.0 m/s) might result from using g=10 m/s2g = 10 \text{ m/s}^2 and a slightly different coefficient. Choice D (29.4 m/s) is too low and doesn't correspond to a reasonable calculation with the given values.

Question 20

A 75 kg person stands on a scale in an elevator that is accelerating upward at 2.5 m/s22.5 \text{ m/s}^2. The elevator cable has a mass of 50 kg and hangs vertically. Air resistance on the elevator and person is negligible, but air resistance on the cable is significant and acts with a force of 300 N downward. What does the scale read?

  1. 735 N735 \text{ N}
  2. 923 N923 \text{ N} (correct answer)
  3. 1035 N1035 \text{ N}
  4. 825 N825 \text{ N}
Explanation: The scale reading equals the normal force the person exerts on the scale, which by Newton's third law equals the normal force the scale exerts on the person. For the person, applying Newton's second law in the upward direction: N−mg=maN - mg = ma, where NN is the normal force from the scale, m=75 kgm = 75 \text{ kg}, and a=2.5 m/s2a = 2.5 \text{ m/s}^2 upward. Therefore: N=mg+ma=75(9.8)+75(2.5)=735+187.5=922.5 NN = mg + ma = 75(9.8) + 75(2.5) = 735 + 187.5 = 922.5 \text{ N}. The information about the cable and air resistance on the cable affects the tension in the elevator cable and the force the elevator motor must provide, but it does not directly affect the force between the person and the scale. The scale reading is 923 N. Choice A (735 N) is just the person's weight mgmg, ignoring the acceleration. Choice C (1035 N) might result from incorrectly adding the air resistance force to the calculation. Choice D (825 N) doesn't correspond to any reasonable calculation.