Newton’s Laws and Free-Body Diagrams (4A) - Chem & Physics

Two people pull on opposite ends of a rope attached to a $10\,\text{kg}$ sled on frictionless ice. Person 1 pulls the sled to the right with $40\,\text{N}$ ; Person 2 pulls to the left with $25\,\text{N}$ . Forces on the sled are the two horizontal tensions, $mg$ , and $N$ . What would be the net force acting on the sled? Take right as positive.
Answer: $15\,\text{N}$ to the right This question tests vector addition of forces to find net force. Newton's Second Law requires summing all forces vectorially to find the net force that determines acceleration. With Person 1 pulling 40 N right and Person 2 pulling 25 N left, the net force is 40 - 25 = 15 N to the right. The correct answer applies simple vector addition with proper sign convention. Answer D incorrectly confuses action-reaction pairs - the two rope tensions act on the same object (the sled) so they don't form an action-reaction pair. When finding net force, add all forces acting on the single object of interest.
A forearm holds a $5.0\,\text{kg}$ mass stationary in the hand with the elbow flexed at $90^\circ$ . Consider the free-body diagram of the hand + held mass treated as one system. External forces on this system include the weight of the mass (downward) and an upward contact force from the forearm at the wrist. Neglect the weight of the hand. Based on Newton's Laws, which statement is most consistent? Assume the system is in static equilibrium.
Answer: The upward wrist contact force equals the weight of the held mass. This question tests force analysis for a composite system in equilibrium. When treating the hand and held mass as one system, external forces must balance for static equilibrium. The system experiences the 5.0 kg mass's weight (49 N) downward and must have an equal upward force from the forearm at the wrist to maintain zero acceleration. The correct answer applies Newton's First Law to the system. Answer B incorrectly claims zero wrist force because muscles act internally - but muscles create the wrist contact force that acts externally on the hand-mass system. For composite systems in equilibrium, external forces must balance.
A $1.2\,\text{kg}$ block is pushed against a horizontal ceiling by an upward applied force $F$ . The block remains at rest, and static friction between the block and ceiling prevents sliding. Forces on the block include applied force (up), weight (down), normal force from ceiling (down), and static friction (horizontal or vertical as needed). Which statement is most consistent with Newton's Laws? Assume the push is purely vertical.
Answer: Static friction can be zero if the applied force balances weight plus the ceiling's normal force. The skill being tested is force balance in unusual orientations like ceiling contact. Newton's first law requires zero net for rest, second for acceleration, third for pairs. The block is at rest against the ceiling with vertical push, so vertical forces balance. Static friction can be zero if no horizontal tendency, with applied balancing weight plus normal, fitting choice D. Choice A is incorrect as friction is horizontal, not vertical here. In such compressed systems, sum vertical forces to zero. Verify friction only if horizontal imbalance exists.
A $7.0\,\text{kg}$ sled is pulled to the right on snow by a horizontal rope tension of $35\,\text{N}$ . The coefficient of kinetic friction between sled and snow is such that the kinetic friction magnitude is $21\,\text{N}$ . Forces on the sled include tension right, friction left, $mg$ down, and $N$ up. What would be the net force acting on the sled? Assume the sled remains in contact with the ground.
Answer: $14\,\text{N}$ to the right This question tests net force calculation with opposing forces. Newton's Second Law requires vector addition of all forces to find net force. With 35 N tension right and 21 N friction left, the net horizontal force is 35 - 21 = 14 N right. The correct answer applies straightforward force subtraction. Answer D incorrectly claims friction always matches pulling force - this only occurs in equilibrium; during acceleration, friction has a fixed magnitude μ_k × N. For net force problems with friction, subtract the friction force from the applied force.
A student stands motionless on a bathroom scale in an elevator that is accelerating upward at $2.0\,\text{m/s}^2$ . The student's mass is $70\,\text{kg}$ . Forces on the student are weight $mg$ downward and the normal force $N$ upward (the scale reading). Based on Newton's Laws, which statement is most consistent? Assume no other forces act.
Answer: The scale reads greater than $mg$ because $N - mg = ma$ upward. This question tests Newton's Second Law applied to accelerating reference frames. When an elevator accelerates upward, objects inside require a net upward force to maintain that acceleration. For the student, applying Newton's Second Law vertically: N - mg = ma (upward positive), which gives N = mg + ma = m(g + a). Since a is positive (upward), the normal force exceeds the weight, causing the scale to read greater than mg. Answer B incorrectly reverses the effect - upward acceleration requires increased normal force, not decreased. When analyzing accelerating systems, always apply F_net = ma in the direction of acceleration.
A $70\,\text{kg}$ person stands still in an elevator that is accelerating upward. The forces on the person are weight $mg$ downward and the normal force $N$ upward. Based on Newton's Laws, which statement is most consistent? Assume the person is not holding onto anything and remains in contact with the floor.
Answer: $N>mg$ because the net force must be upward. The skill being tested is determining apparent weight in accelerating elevators. Newton's first law requires balanced forces for constant velocity, the second links net force to acceleration, and the third pairs forces. The elevator accelerates upward, so net force on the person is upward, meaning normal exceeds weight. N > mg because it provides the upward net force for acceleration, aligning with choice C. Choice B is incorrect as N equals mg only for zero acceleration, not here. For related questions, compare normal force to weight based on acceleration direction. Check by noting N - mg = m a for upward acceleration.
A $6.0\,\text{kg}$ box is at rest on a horizontal surface. A horizontal force is applied and gradually increased. The maximum static friction force available is $18\,\text{N}$ . When the applied force reaches $12\,\text{N}$ , the box still does not move. Which statement is most consistent with Newton's Laws and static friction? Forces: applied (right), static friction (left), weight, normal.
Answer: Static friction is $12\,\text{N}$ opposing the applied force. The skill being tested is understanding static friction in pre-motion scenarios. Newton's first law maintains rest with zero net force, the second links to acceleration, and the third pairs. The box remains at rest, so static friction balances the applied force. Friction is 12 N opposing, less than maximum, fitting choice B. Choice C is incorrect because friction is nonzero to counter the applied force. For friction problems, note static friction adjusts to match applied up to max. Verify by checking if applied < max friction implies no motion.
A $0.20\,\text{kg}$ ball is thrown straight upward. Neglect air resistance. After it leaves the hand and before it reaches the top, which statement about the net force on the ball is most consistent with Newton's Laws? Forces: gravity only (downward).
Answer: Net force is downward with magnitude $mg$ . The skill being tested is net force in projectile motion neglecting air. Newton's first law maintains velocity without net force, but gravity provides net force per second law, with third pairing. The ball is in free fall upward, with only gravity acting downward. Net force is mg downward constantly, supporting choice C. Choice A is incorrect as net force direction opposes motion when rising. In projectile problems, note gravity is the only force post-throw. Verify by recalling acceleration is constant g downward.
A $10\,\text{kg}$ sled is pulled across snow by a horizontal rope tension of $40\,\text{N}$ . The coefficient of kinetic friction is such that the friction force magnitude is $25\,\text{N}$ . What would be the net force acting on the sled? Forces: tension (forward), kinetic friction (backward), weight, normal.
Answer: $15\,\text{N}$ forward The skill being tested is calculating net force from tension and friction. Newton's first law needs zero net for constant velocity, but here implied acceleration from unbalanced forces per second law, with third law pairing. The sled has forward tension and backward friction, so net force is their difference. Net force is 40 N - 25 N = 15 N forward, logically choice A. Choice D is incorrect as vertical cancellation does not affect horizontal net. For such problems, subtract opposing horizontal forces. Verify by predicting acceleration if net is nonzero.
A $4.0\,\text{kg}$ block rests on a rough incline at $25^\circ$ above horizontal and does not slide. Forces on the block are weight $mg$ downward, normal force $N$ perpendicular to the surface, and static friction $f_s$ parallel to the surface. Based on Newton's Laws, which statement about the direction of static friction is most consistent?
Answer: Static friction points up the incline, opposing the tendency of the block to slide down. This question tests understanding static friction's role in preventing motion on inclined planes. Newton's first law requires zero net force for the stationary block, meaning forces parallel to the incline must balance. The weight component parallel to the incline is mg sin 25° pointing down the slope, creating a tendency to slide downward. Static friction must oppose this tendency by pointing up the incline with magnitude $f_s$ = mg sin 25°, preventing any sliding. Choice B incorrectly claims friction acts downward with gravity, while choice C incorrectly assumes zero friction for stationary objects. Choice D confuses friction's direction with the normal force direction. The key principle is that static friction always opposes the direction of impending motion - on an incline, gravity tends to cause downward sliding, so friction must point upward to maintain equilibrium.
A car of mass $m$ travels around a flat (unbanked) circular track at constant speed. The tires provide static friction with the road. Forces on the car are $mg$ downward, normal force $N$ upward, and static friction horizontally toward the center of the circle. Based on Newton's Laws, which statement is most consistent? Assume no slipping occurs.
Answer: Static friction provides the centripetal net force toward the center. This question tests understanding of circular motion and centripetal force. Newton's Second Law for circular motion requires a net force toward the center to maintain the centripetal acceleration v²/r. On a flat track, only friction can provide a horizontal force, so static friction must point toward the center to keep the car on its circular path. The correct answer identifies friction as providing the required centripetal force. Answer B incorrectly claims zero net force - constant speed doesn't mean zero acceleration in circular motion, as direction constantly changes. For horizontal circular motion, friction provides the centripetal force.
A $3.0\,\text{kg}$ cart on a frictionless horizontal track is pushed by a constant horizontal force of $9.0\,\text{N}$ to the right. The cart starts from rest. Forces on the cart are the applied force, weight $mg$ , and normal force $N$ . What would be the net force acting on the cart? Assume the vertical forces balance.
Answer: $9.0\,\text{N}$ to the right This question tests application of Newton's Second Law to determine net force. Newton's Second Law states that net force equals mass times acceleration (F_net = ma), and for objects starting from rest, the net force determines the direction of acceleration. Since the cart experiences only a 9.0 N horizontal force to the right (with vertical forces balanced), this becomes the net force. The correct answer directly applies F_net = F_applied when no opposing horizontal forces exist. Answer B is incorrect because starting from rest doesn't mean zero net force - it means zero initial velocity. For problems with single unbalanced forces, the net force equals that unbalanced force.
A $6.0\,\text{kg}$ block is pressed against a vertical wall by a horizontal force of $120\,\text{N}$ to the right. The block remains at rest; static friction between block and wall prevents sliding. Forces on the block are $mg$ downward, normal force from the wall to the left, the applied push to the right, and static friction upward. Based on Newton's Laws, which statement is most consistent? Assume equilibrium in both directions.
Answer: Static friction must balance $mg$ in the vertical direction. This question tests force balance in two perpendicular directions. For the block to remain at rest against the wall, forces must balance both horizontally and vertically. Horizontally: the 120 N push right balances the normal force left. Vertically: with no other vertical forces, static friction must balance the weight mg to prevent sliding. The correct answer identifies this vertical force balance requirement. Answer A incorrectly states friction opposes the push - friction opposes relative motion, which here would be vertical sliding due to gravity. For objects pressed against vertical surfaces, friction balances weight while normal force balances the push.
An elevator car of mass $m$ is moving upward at constant velocity. Forces on the car are the cable tension $T$ upward and the weight $mg$ downward (air resistance is negligible). Based on Newton's Laws, which statement about the system is most consistent?
Answer: $T=mg$ , because constant velocity implies zero net force. This question tests the application of Newton's first law to constant velocity motion. Newton's first law states that an object moving at constant velocity has zero net force, meaning all forces must balance. For the elevator moving upward at constant velocity, the acceleration is zero, so by F_net = ma, the net force must also be zero. This requires the upward cable tension T to exactly equal the downward weight mg, giving T = mg. Choice A incorrectly assumes upward motion requires net upward force, confusing velocity with acceleration. Choice B incorrectly suggests gravity must dominate during upward motion, while choice D confuses position with force. The key insight is that constant velocity (regardless of direction) always means zero acceleration and therefore zero net force - motion doesn't require force to maintain, only to change.
A $1.0\,\text{kg}$ sled slides to the right on a horizontal surface. A student applies a horizontal force of $5.0\,\text{N}$ to the left. Kinetic friction has magnitude $2.0\,\text{N}$ and acts opposite the sled's motion. Vertical forces are $N$ upward and $mg$ downward. Taking right as positive, what would be the net force acting on the sled?
Answer: $-7.0\,\text{N}$ , because both the applied force and friction act to the left. This question tests vector addition of forces with careful attention to direction conventions. Newton's second law states that net force equals the vector sum of all forces, with sign indicating direction. Taking right as positive, the sled moving right experiences kinetic friction of 2.0 N to the left (negative direction), giving -2.0 N. The applied force of 5.0 N to the left also contributes -5.0 N. The net horizontal force is (-5.0 N) + (-2.0 N) = -7.0 N, indicating 7.0 N to the left. Choice A (+3.0 N) incorrectly subtracts the magnitudes, choice C (-3.0 N) incorrectly treats friction as opposing the applied force rather than the motion, and choice D (0 N) ignores horizontal forces entirely. Remember that kinetic friction always opposes the direction of motion, not necessarily the applied force - here both forces act leftward, combining rather than canceling.
A $12\,\text{kg}$ equipment crate is pulled across a horizontal lab floor at constant velocity by a rope that makes a $30^\circ$ angle above the horizontal. The rope tension is $T=80\,\text{N}$ . Kinetic friction acts between the crate and floor. Neglect air resistance. Based on Newton's Laws, which statement about the horizontal forces is most consistent with the motion? (Forces on crate: weight $mg$ downward, normal force $N$ upward, tension along rope, kinetic friction opposite motion.)
Answer: The kinetic friction force equals the horizontal component of tension. This question tests understanding of Newton's First Law and equilibrium conditions for objects moving at constant velocity. Newton's First Law states that an object maintains constant velocity when the net force is zero, requiring all forces to be balanced. For the crate moving at constant velocity, the horizontal forces must sum to zero: the horizontal component of tension (T cos 30°) must equal the kinetic friction force opposing motion. The correct answer follows from applying equilibrium conditions to the horizontal direction. Answer A is incorrect because if friction exceeded the horizontal tension component, the crate would decelerate rather than maintain constant velocity. When analyzing constant velocity problems, always check that net force equals zero in all directions.
A small drone of mass $m$ hovers motionless. The thrust force from the rotors acts upward; weight acts downward. Wind is negligible. Based on Newton's Laws, which statement is most consistent? Consider only vertical forces on the drone.
Answer: Thrust equals weight, giving zero net force. The skill being tested is equilibrium in hovering objects. Newton's first law states zero net force for constant position, the second equates force to m a, and the third pairs. The drone hovers motionless, so net vertical force is zero. Thrust equals weight for balance, making choice B correct. Choice A is incorrect as thrust equals, not exceeds, weight in hover. In vertical equilibrium cases, set upward forces equal to downward. Verify by confirming no acceleration implies balanced forces.
A $9.0\,\text{kg}$ suitcase is pulled across an airport floor by a strap at $20^\circ$ above horizontal. The suitcase accelerates to the right. Forces include weight, normal, strap tension, and kinetic friction. Based on Newton's Laws, which statement is most consistent? Assume friction acts opposite motion and air resistance is negligible.
Answer: The net horizontal force must be to the right. The skill being tested is identifying net force direction in accelerated motion with inclined pull. Newton's first law for constant motion, second for acceleration from net force, third for pairs. The suitcase accelerates right, so net horizontal force is to the right. This follows from unbalanced forward components over friction, making choice A correct. Choice B is incorrect as net force is not zero during acceleration. For inclined force problems, resolve into components. Check if net horizontal >0 matches acceleration direction.
A $2.0\,\text{kg}$ cart on a frictionless track is pulled to the right by a constant horizontal force of $10\,\text{N}$ . A second force of $4\,\text{N}$ acts to the left due to a spring attached to the cart. The cart is observed to accelerate. What is the net force acting on the cart? Forces on the cart: applied pull (right), spring force (left), weight (down), normal force (up).
Answer: $6\,\text{N}$ to the right The skill being tested is identifying net force from multiple forces using free-body diagrams. Newton's first law indicates no net force for constant velocity, the second law states net force equals mass times acceleration, and the third involves equal-opposite reactions. Here, the cart accelerates due to unbalanced horizontal forces, so Newton's second law applies with net force causing acceleration. The net force is 10 N right minus 4 N left, equaling 6 N to the right, logically following as choice A. Choice D is incorrect because vertical forces canceling does not mean horizontal net force is zero. In similar problems, sum all forces vectorially to find the net force. Verify by checking if the net force aligns with observed acceleration direction.
A $0.50\,\text{kg}$ puck moves in a straight line on a horizontal air table with negligible friction. A constant horizontal force of $2.0\,\text{N}$ is applied to the left. Forces on the puck include the applied force, $mg$ , and $N$ . Based on Newton's Laws, which statement about the puck's motion is most consistent? Assume the puck initially moves to the right.
Answer: The puck accelerates to the left regardless of its initial direction of motion. This question tests Newton's Second Law with constant force application. Newton's Second Law states that acceleration depends on net force, not on velocity: a = F_net/m. With a 2.0 N force applied left on a 0.50 kg puck, acceleration is 2.0/0.50 = 4.0 m/s² to the left, regardless of initial motion direction. The correct answer recognizes that force determines acceleration direction, not velocity. Answer A incorrectly assumes zero friction means constant velocity - zero friction means no friction force, but other forces can still cause acceleration. For constant force problems, acceleration direction matches net force direction, independent of velocity.

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