A sign hangs at rest from two identical vertical cords. The forces on the sign are: weight downward (↓) and two upward tensions and (↑). Take .
What is the tension in each cord?
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Physics Quiz
Practice Explain Force Magnitude And Direction in Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A 2.0kg sign hangs at rest from two identical vertical cords. The forces on the sign are: weight Fg downward (↓) and two upward tensions FT1 and FT2 (↑). Take g=10m/s2.
What is the tension in each cord?
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A 2.0kg sign hangs at rest from two identical vertical cords. The forces on the sign are: weight Fg downward (↓) and two upward tensions FT1 and FT2 (↑). Take g=10m/s2.
What is the tension in each cord?
Explanation: This question tests understanding of force balance and equilibrium with multiple tensions. An object is in equilibrium when the net force is zero, meaning the vector sum of all forces equals zero—for an object at rest or moving at constant velocity, forces must balance in every direction (sum of upward forces = sum of downward forces, sum of rightward forces = sum of leftward forces). In this scenario, the sign is at rest, so forces must balance. Vertically: the two identical tensions (F_{T1} and F_{T2}) upward must equal weight downward (F_g = 2 kg * 10 m/s² = 20 N), so each tension = 10 N since they are equal. Choice A is correct because it properly applies equilibrium conditions showing forces balance vertically, with each cord supporting half the weight. Choice B incorrectly doubles the tension or assumes only one cord, but with two identical cords, the weight is shared equally. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A 5.0kg box rests on a 30∘ incline. The coefficient of kinetic friction is μk=0.20, and the box is sliding down the incline. Take g=10m/s2.
Which direction does the kinetic friction force Ff on the box act?
Explanation: This question tests understanding of force magnitude and direction in multi-force scenarios on inclines. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, F_x = F cos(θ) and F_y = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). On the incline, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (5 kg)(10 m/s²) sin(30°) = 25 N down the slope, and perpendicular is F_perp = mg cos(θ) = 43.3 N into the surface. The normal force equals the perpendicular component: F_N ≈ 43.3 N, and since the box is sliding down, kinetic friction acts up the slope to oppose the motion. Choice B is correct because it identifies the force relationship from given constraints, with friction opposing the direction of sliding (down the incline), so it points up the incline. Choice A shows friction pointing in the wrong direction, but friction always opposes motion or the tendency to move, so it must point up the incline to oppose the downhill sliding. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, F_N = F_g and F_app = F_f; on incline at angle θ, F_N = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A 10kg sled is pulled on level snow by a horizontal rope force Fapp=80N to the right (→). The coefficient of kinetic friction is μk=0.30. Take g=10m/s2. Forces: Fg (↓), FN (↑), Fapp (→), Ff (←).
What is the net horizontal force on the sled (magnitude and direction)?
Explanation: This question tests understanding of net force determination in multi-force scenarios. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the sled are: weight 100 N downward, normal force 100 N upward, applied force 80 N to the right, and friction F_f = μ F_N = 0.30 * 100 = 30 N to the left. The net vertical force is 0 N (balanced). The net horizontal force is 80 - 30 = 50 N to the right. Choice C is correct because it correctly calculates net force as vector sum with directions, considering opposing forces subtract. Choice D calculates an individual force magnitude when the question asks for net force—net force requires vector addition considering directions: F_net = F_right - F_left, not just the magnitude of one force. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A 6.0kg box rests on a 25∘ incline with no other forces applied. Take g=10m/s2. Consider the component of the weight parallel to the incline.
What is the magnitude of the component of the gravitational force parallel to the incline, Fg,∥=mgsinθ?
Explanation: This question tests understanding of component analysis of angled forces on inclines. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, F_x = F cos(θ) and F_y = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). On the incline, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (6 kg)(10 m/s²) sin(25°) ≈ 60 * 0.4226 ≈ 25 N down the slope, and perpendicular is F_perp = mg cos(θ) ≈ 54 N into the surface. Choice A is correct because it accurately resolves the gravitational force into components using correct trigonometry, with F_{g,parallel} = mg sinθ ≈ 25 N. Choice B uses sine when it should use cosine (or vice versa) for the component calculation—for the parallel component on an incline, it's mg sin(θ), not mg cos(θ). When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, F_N = F_g and F_app = F_f; on incline at angle θ, F_N = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A 20kg crate is pulled across a horizontal floor by a rope that makes a 30∘ angle above the horizontal. The rope tension is FT=100N. Take g=10m/s2.
What is the vertical component of the tension force on the crate, FT,y?
Explanation: This question tests understanding of component analysis of angled forces. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, F_x = F cos(θ) and F_y = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). The tension force of 100 N at 30° above horizontal has components: horizontal F_x = 100 cos(30°) ≈ 87 N rightward, and vertical F_y = 100 sin(30°) = 50 N upward, which would affect the normal force if considered. Choice B is correct because it accurately resolves force into components using correct trigonometry, with F_{T,y} = F_T sinθ = 50 N. Choice A uses cosine for the vertical component when it should use sine—for a force at angle θ from horizontal, the vertical component uses sine: F_y = F sin(θ), not F cos(θ). When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A 10kg box is pulled across a horizontal floor by an applied force Fapp=60N at 30∘ above the horizontal (to the right). Assume g=10m/s2. Which expression best gives the normal force magnitude FN on the box (assuming the box does not accelerate vertically)?
Explanation: This question tests understanding of relationships between force types. The normal force is the contact force exerted by a surface perpendicular to the surface, with magnitude that adjusts to prevent the object from passing through the surface—it's not always equal to the object's weight and depends on other forces and surface orientation. The applied force of F_app = 60 N at angle 30° above horizontal has components: horizontal F_x = F_app cos(30°) rightward, and vertical F_y = F_app sin(30°) upward, which affects the normal force making F_N = mg - F_y = mg - F_app sin(30°), since no vertical acceleration. Vertically, forces balance: F_N + F_app sin(30°) = mg. Choice B is correct because it correctly determines force magnitude from force balance or given constraints. Choice D assumes the normal force always equals the weight (F_N = mg), but this is only true on horizontal surfaces with no other vertical forces—here the upward component of the applied force reduces the normal force to F_N = mg - F_app sin(30°). When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A 18kg crate is pulled to the right (→) across a horizontal floor with an applied force Fapp=90N. The coefficient of kinetic friction is μk=0.25 and g=10m/s2. What is the magnitude of the net force on the crate?
Explanation: This question tests understanding of net force determination. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the crate are: weight downward F_g = (18 kg)(10 m/s²) = 180 N, normal force upward F_N = 180 N (balances vertically), applied force 90 N right, and kinetic friction F_f = μ_k F_N = 0.25 * 180 = 45 N left. The net horizontal force is 90 N - 45 N = 45 N to the right, so the crate accelerates rightward. Choice B is correct because it correctly calculates net force as vector sum with directions. Choice A incorrectly assumes the object is in equilibrium (forces balance), when actually the object is accelerating so there must be a net force in the direction of acceleration. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, F_N = F_g and F_app = F_f; on incline at angle θ, F_N = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A 10kg box is on a horizontal floor. A student pushes straight downward on the top of the box with an additional force of 40N (↓). The box is at rest (no vertical acceleration). Take g=10m/s2. What is the magnitude of the normal force FN exerted by the floor on the box?
Explanation: This question tests understanding of normal force determination. The normal force is the contact force exerted by a surface perpendicular to the surface, with magnitude that adjusts to prevent the object from passing through the surface—it's not always equal to the object's weight and depends on other forces and surface orientation. In this scenario, the box is at rest with an additional downward push of 40 N, so vertical forces: weight F_g = (10 kg)(10 m/s²) = 100 N downward, push 40 N downward, and normal force F_N upward; since no vertical acceleration, F_N = 100 N + 40 N = 140 N. Choice C is correct because it correctly determines force magnitude from force balance or given constraints. Choice A assumes the normal force equals the weight (F_N = mg = 100 N), but this is only true with no other vertical forces—here the additional downward push increases the normal force to F_N = mg + push. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A 5kg block is pulled along a horizontal surface by a force Fapp=40N to the right (→). Kinetic friction is Ff=10N to the left (←). Weight and normal force act vertically and cancel. Using g=10m/s2 if needed, what is the magnitude of the net force on the block, and in which direction does it act?
Explanation: This question tests understanding of net force determination. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the object are: weight downward, normal force upward (cancel vertically), applied force 40 N to the right, and friction 10 N to the left. The net vertical force is 0 N (balanced). The net horizontal force is F_app - F_f = 40 - 10 = 30 N to the right, so the object accelerates rightward. Choice A is correct because it correctly calculates net force as vector sum with directions. Choice D incorrectly assumes the object is in equilibrium (forces balance), when actually the object is accelerating so there must be a net force in the direction of acceleration. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, F_N = F_g and F_app = F_f; on incline at angle θ, F_N = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A 25kg sign hangs at rest from a single vertical rope. Use g=10m/s2. The forces on the sign are tension FT upward (↑) and weight Fg downward (↓). What is the magnitude of the rope tension FT?
Explanation: This question tests understanding of force balance and equilibrium. An object is in equilibrium when the net force is zero, meaning the vector sum of all forces equals zero—for an object at rest or moving at constant velocity, forces must balance in every direction (sum of upward forces = sum of downward forces, sum of rightward forces = sum of leftward forces). In this scenario, the sign is at rest, so forces must balance. Vertically: tension upward (F_T) must equal weight downward (F_g = mg = (25 kg)(10 m/s²) = 250 N), giving F_T = 250 N. Choice B is correct because it properly applies equilibrium conditions showing forces balance. Choice A incorrectly calculates tension as mass without gravity (25 N), but tension must balance the full weight mg. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Common errors to avoid: (a) assuming normal force always equals weight (it depends on other vertical forces and surface angle), (b) forgetting that friction opposes motion or tendency to move (not just opposes applied force), (c) confusing action-reaction pairs (different objects) with balanced forces (same object), and (d) forgetting to account for all forces when finding net force.
A 6kg cart moves to the right on a horizontal track with acceleration a=2m/s2. A motor applies Fapp=20N to the right (→), and friction is Ff=8N to the left (←). Weight and normal cancel vertically. Based on Newton's 2nd law, what is the magnitude of the net force on the cart?
Explanation: This question tests understanding of net force determination. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the cart are: applied force 20 N to the right, friction 8 N to the left, weight and normal canceling vertically. The net horizontal force is 20 N - 8 N = 12 N to the right, and from Newton's Second Law, F_net = ma = (6 kg)(2 m/s²) = 12 N, confirming consistency. Choice B is correct because it correctly calculates net force as vector sum with directions. Choice C calculates an individual force magnitude when the question asks for net force—net force requires vector addition considering directions: F_net = F_right - F_left, not just the magnitude of one force. When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, F_N = F_g and F_app = F_f; on incline at angle θ, F_N = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).
A sign of weight 200N hangs at rest from two identical cables. Each cable makes a 60∘ angle with the horizontal ceiling.
What is the tension magnitude FT in each cable?
Explanation: This question tests understanding of force balance and equilibrium in a symmetric cable support system. An object is in equilibrium when the net force is zero, meaning the vector sum of all forces equals zero—for an object at rest, forces must balance in every direction. For the hanging sign, three forces act: weight 200 N downward and two equal cable tensions F_T at 60° from horizontal. Since the sign is at rest, the vertical components of both cable tensions must sum to equal the weight. Each cable's vertical component is F_T sin(60°) = F_T × 0.866. For vertical equilibrium: 2 × F_T sin(60°) = 200 N, so 2 × F_T × 0.866 = 200 N, giving F_T = 200 N / (2 × 0.866) = 200 N / 1.732 = 115.5 N ≈ 115 N. Choice B (115 N) is correct because it properly applies equilibrium conditions, recognizing that two cables share the load and using the sine of the angle from horizontal for the vertical component. Choice A (100 N) incorrectly assumes each cable carries half the weight without accounting for the angle, while Choice C (200 N) wrongly assumes each cable carries the full weight. When analyzing symmetric support systems: (1) identify that multiple supports share the load, (2) resolve each support force into components, (3) sum all vertical components to equal the supported weight, and (4) for cables at angle θ from horizontal, use sin(θ) for vertical components. Remember that steeper cables (larger angle from horizontal) have larger vertical components, requiring less tension to support the same weight.
A sled of mass 25kg is pulled on level snow by a horizontal rope force FT=120N to the right (→). The kinetic friction force has magnitude Ff=45N to the left (←). Take g=10m/s2.
What is the magnitude of the net force on the sled?
Explanation: This question tests understanding of net force determination when multiple horizontal forces act on an object. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when forces don't balance, the object accelerates in the direction of the net force with magnitude a = F_net / m. The forces acting on the sled are: tension force F_T = 120 N to the right (→) and friction force F_f = 45 N to the left (←). Since these forces act in opposite directions, we must subtract them to find the net force. The net horizontal force is F_net = F_T - F_f = 120 N - 45 N = 75 N to the right, so the sled accelerates rightward with net force magnitude 75 N. Choice C (75 N) is correct because it properly calculates net force as the vector sum, subtracting the opposing friction force from the applied tension force. Choice B (45 N) shows only the friction force magnitude, ignoring the tension force, while Choice D (165 N) incorrectly adds the forces instead of subtracting them—forces in opposite directions must subtract, not add. When calculating net force: (1) identify all forces acting on the object and their directions, (2) forces in the same direction add while forces in opposite directions subtract, (3) the net force direction is toward whichever side has the larger total force, and (4) use F_net = ma to find acceleration if needed. Common errors include forgetting to account for direction (treating all forces as positive) or calculating individual force magnitudes when the question asks for net force.
A 6kg box is at rest on a horizontal surface. A horizontal pull of Fapp=25N acts to the right (→). The coefficient of static friction is μs=0.50. Take g=10m/s2.
What is the magnitude of the static friction force Ff on the box?
Explanation: This question tests understanding of static friction and equilibrium for an object at rest. An object at rest has zero acceleration, which means the net force must be zero—all forces must balance in every direction, with static friction adjusting to prevent motion. In this scenario, the box remains at rest despite the applied force, so horizontal forces must balance. The maximum possible static friction is F_f,max = μ_s × F_N = 0.50 × (6 kg × 10 m/s²) = 0.50 × 60 N = 30 N. Since the applied force (25 N) is less than this maximum (30 N), the box doesn't slide, and static friction exactly balances the applied force: F_f = 25 N to the left (←) to oppose the 25 N rightward pull. Choice B (25 N) is correct because static friction adjusts to exactly balance the applied force when the object remains at rest—it's not automatically at its maximum value. Choice C (30 N) incorrectly assumes static friction is always at its maximum value μ_s × F_N, but static friction only reaches this maximum at the threshold of motion—below this threshold, it equals whatever is needed to maintain equilibrium. When analyzing static friction: (1) calculate the maximum possible static friction F_f,max = μ_s × F_N, (2) compare the applied force to this maximum, (3) if F_app < F_f,max, the object remains at rest and F_f = F_app, (4) if F_app ≥ F_f,max, the object begins to slide and kinetic friction takes over. Key insight: static friction is a responsive force that adjusts between 0 and μ_s × F_N to prevent motion.
A 10kg box rests on a 30∘ incline (angle measured from the horizontal). The coefficient of kinetic friction is μk=0.20, and the box slides down the incline. Using g=10m/s2, what is the magnitude of the normal force FN on the box?
Explanation: This question tests understanding of component analysis of angled forces on an inclined plane. When an object rests on an incline, the weight force must be resolved into components: parallel to the incline and perpendicular to the incline—the normal force is the contact force exerted by the surface perpendicular to the surface, equal in magnitude to the perpendicular component of weight when no other forces act perpendicular to the surface. On the incline at 30°, we resolve the weight into components: perpendicular to incline is F_perp = mg cos(θ) = (10 kg)(10 m/s²) cos(30°) = 100 N × 0.866 = 86.6 N into the surface, and parallel to incline is F_parallel = mg sin(θ) = 100 N × 0.5 = 50 N down the slope. The normal force equals the perpendicular component: F_N = 86.6 N, since there are no other forces acting perpendicular to the incline surface. Choice B is correct because it accurately resolves the weight force into components using correct trigonometry—the perpendicular component uses cosine of the incline angle. Choice C incorrectly assumes the normal force equals the full weight (F_N = mg = 100 N), but this is only true on horizontal surfaces—on an incline, the normal force equals only the perpendicular component of weight, which is mg cos(θ). When analyzing forces on inclines: (1) always resolve weight into components parallel and perpendicular to the surface, (2) use F_perp = mg cos(θ) and F_parallel = mg sin(θ) where θ is measured from horizontal, (3) remember the normal force equals the perpendicular component of weight (not the full weight), and (4) friction acts parallel to the surface opposing motion or tendency to move.
A skydiver of mass 7kg is falling straight downward at terminal velocity. The drag force FD acts upward. Using g=10m/s2, what is the magnitude of the drag force FD?
Explanation: This question tests understanding of force balance at terminal velocity. Terminal velocity occurs when a falling object reaches constant velocity—at this point, the net force is zero because the upward drag force exactly balances the downward weight force, resulting in equilibrium despite motion. In this scenario, the skydiver falls at terminal velocity (constant velocity), so forces must balance vertically. The forces acting are: weight F_g = mg = (7 kg)(10 m/s²) = 70 N downward, and drag force F_D upward. Since terminal velocity means constant velocity (zero acceleration), the net force must be zero: F_D = F_g = 70 N upward to balance the weight. Choice B is correct because it properly applies equilibrium conditions at terminal velocity—the drag force must equal the weight for the net force to be zero. Choice C incorrectly suggests zero drag force, but this would result in free fall acceleration, not terminal velocity—terminal velocity requires drag to balance weight, not be zero. When analyzing terminal velocity: (1) recognize that terminal velocity means constant velocity with zero acceleration, (2) apply Newton's First Law requiring zero net force for constant velocity motion, (3) identify that drag must exactly balance weight at terminal velocity, and (4) remember that equilibrium can occur during motion, not just at rest.
A 15kg box rests on a 20∘ incline (angle from the horizontal) and is held at rest by static friction only (no rope). Using g=10m/s2, what is the required magnitude of the static friction force Ff (along the incline) to keep the box in equilibrium?
Explanation: This question tests understanding of force balance on an inclined plane in static equilibrium. An object is in equilibrium when the net force is zero—for an object at rest on an incline held by friction alone, the static friction force must exactly balance the component of weight parallel to the incline. On the incline at 20°, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (15 kg)(10 m/s²) sin(20°) = 150 N × 0.342 = 51.3 N down the slope, and perpendicular is F_perp = mg cos(θ) = 150 N × 0.940 = 141 N into the surface. Since the box is at rest (equilibrium), the static friction force up the incline must equal the parallel component of weight: F_f = 51.3 N ≈ 51 N up the slope. Choice C is correct because it accurately calculates the parallel component of weight that friction must balance for equilibrium on the incline. Choice B incorrectly identifies the perpendicular component (141 N) as the friction force, but friction acts parallel to the surface, not perpendicular—the perpendicular component determines the normal force, not friction. When analyzing static equilibrium on inclines: (1) resolve weight into components parallel and perpendicular to the surface, (2) recognize that static friction adjusts to balance the parallel component of weight (up to its maximum value), (3) use F_parallel = mg sin(θ) for the component friction must balance, and (4) remember friction acts parallel to the surface, opposing the tendency to slide.
A 5kg sled is pulled on level snow by a rope with tension FT=40N at 30∘ above the horizontal. The kinetic friction force has magnitude Ff=10N opposing the motion. Using g=10m/s2, what is the net force in the horizontal direction (magnitude and direction)?
Explanation: This question tests understanding of net force determination when forces act at angles. Newton's Second Law (F_net = ma) states that the net force on an object equals its mass times acceleration—when calculating net force, we must consider both magnitude and direction of all forces, resolving angled forces into components. The applied force of F_T = 40 N at angle 30° above horizontal has components: horizontal F_x = F_T cos(30°) = 40 N × 0.866 = 34.6 N rightward, and vertical F_y = F_T sin(30°) = 40 N × 0.5 = 20 N upward. The forces acting horizontally are: tension component 34.6 N to the right and friction 10 N to the left. The net horizontal force is F_net = 34.6 N - 10 N = 24.6 N to the right, so the sled accelerates rightward. Choice B is correct because it properly resolves the angled force into components and correctly calculates net force as the vector sum considering directions. Choice D uses the horizontal component of tension (34.6 N) but forgets to subtract the opposing friction force—net force requires vector addition considering directions: F_net = F_right - F_left, not just the magnitude of one force. When analyzing net forces with angled components: (1) resolve all angled forces into x and y components using trigonometry, (2) find net force in each direction separately by adding forces in same direction and subtracting opposing forces, (3) remember that F_x = F cos(θ) for angle from horizontal, and (4) always specify both magnitude and direction for net force vectors.
A 6kg crate is pulled to the right across a horizontal floor at constant velocity by a horizontal applied force Fapp=30N. What is the magnitude of the kinetic friction force Ff acting on the crate?
Explanation: This question tests understanding of force balance and equilibrium for an object moving at constant velocity. An object moving at constant velocity has zero acceleration, which means the net force must be zero according to Newton's First Law—this requires all forces to balance in every direction. In this scenario, the crate moves horizontally at constant velocity, so forces must balance both vertically and horizontally. Vertically: normal force upward (F_N) equals weight downward (F_g = mg = 6 kg × 10 m/s² = 60 N), so forces balance. Horizontally: the applied force F_app = 30 N to the right must be balanced by an equal friction force to the left, so F_f = 30 N opposing the motion. Choice B is correct because it properly applies equilibrium conditions showing that when an object moves at constant velocity, the friction force must exactly balance the applied force. When analyzing forces on objects at constant velocity: (1) recognize that constant velocity means zero acceleration and therefore zero net force, (2) identify all forces and their directions, (3) apply force balance in each direction separately, and (4) remember that kinetic friction opposes the direction of motion with magnitude that depends on the normal force and coefficient of friction, but in equilibrium problems it must equal the applied force.
A 20kg box rests on a frictionless 30∘ incline (angle measured from the horizontal). Forces on the box: weight Fg (↓), normal force FN (perpendicular to the incline), and no friction. Using g=10m/s2, what is the magnitude of the normal force FN?
Explanation: This question tests understanding of component analysis of angled forces. When forces act at angles, they must be resolved into perpendicular components using trigonometry: for a force F at angle θ from the horizontal, F_x = F cos(θ) and F_y = F sin(θ), or on an incline the weight component parallel to slope is mg sin(θ) and perpendicular is mg cos(θ). On the incline, we resolve the weight into components: parallel to incline is F_parallel = mg sin(θ) = (20 kg)(10 m/s²) sin(30°) = 200 * 0.5 = 100 N down the slope, and perpendicular is F_perp = mg cos(θ) = 200 cos(30°) = 200 * (√3/2) ≈ 173 N into the surface. The normal force equals the perpendicular component: F_N = 173 N, and since it's frictionless, no friction balances the parallel component, but the question asks for F_N. Choice B is correct because it accurately resolves force into components using correct trigonometry, F_N = mg cos(θ). Choice C assumes the normal force equals the weight (F_N = mg = 200 N), but this is only true on horizontal surfaces with no other vertical forces—here the incline changes the normal force to F_N = mg cos(θ). When analyzing forces: (1) draw a free body diagram showing all forces on the object, (2) choose a coordinate system and resolve angled forces into components, (3) apply equilibrium conditions (ΣF = 0 in each direction) if object is at rest or constant velocity, or apply F_net = ma if accelerating, and (4) remember that forces in opposite directions subtract while forces in same direction add. Key relationships to remember: on horizontal surface in equilibrium, F_N = F_g and F_app = F_f; on incline at angle θ, F_N = mg cos(θ) and component down slope = mg sin(θ); for constant velocity motion (net force = 0 in all directions); for accelerating motion (net force ≠ 0 and F_net = ma points in direction of acceleration).