MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4a Newtons Laws Free Body

What this quiz covers

This quiz focuses on 4a Newtons Laws Free Body, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

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.

Question 1

1. The normal force on the top block equals the weight of both blocks combined.
2. The normal force on the top block equals the weight of the top block. (correct answer)
3. The normal force on the top block is zero because it is not touching the table.
4. The net force on the top block is downward because gravity is unbalanced.

Explanation: This question tests force analysis on individual objects in a stacked system. When analyzing only the top block, consider forces acting directly on it: its own weight (3.0 kg × g) downward and the normal force from the bottom block upward. For equilibrium, these must balance, giving N = mg for the top block alone. The correct answer applies Newton's First Law to the isolated top block. Answer A incorrectly includes the bottom block's weight - this acts on the bottom block, not the top block. When drawing free-body diagrams, include only forces acting on the specific object being analyzed.

Question 2

1. The tension equals the weight of the hanging mass because the string is light.
2. The tension is less than the weight of the hanging mass because that mass accelerates downward. (correct answer)
3. The tension is greater than the weight of the hanging mass because the cart accelerates.
4. The cart has no horizontal net force because $N$ cancels $T$.

Explanation: This question tests Newton's Second Law for connected objects with common acceleration. When objects are connected by a light string over a frictionless pulley, they share the same acceleration magnitude. For the hanging mass accelerating downward: mg - T = ma, giving T = m(g - a). Since acceleration is non-zero and downward for the hanging mass, tension must be less than its weight. The correct answer recognizes this constraint from Newton's Second Law. Answer A incorrectly assumes tension equals weight - this only occurs in equilibrium, not during acceleration. For accelerating connected systems, tension differs from weight by ma.

Question 3

1. $T = mg$ and the net force is zero. (correct answer)
2. $T > mg$ because tension must exceed weight to keep the mass from falling.
3. $T < mg$ because the spring supports part of the weight.
4. The net force is $mg$ downward but acceleration is zero because velocity is zero.

Explanation: This question tests Newton's First Law for static equilibrium. When an object hangs at rest, it has zero acceleration, requiring the net force to be zero according to F_net = ma. For the hanging mass, only two forces act: weight mg downward and tension T upward. These must be equal in magnitude for equilibrium, giving T = mg. The correct answer applies the equilibrium condition directly. Answer B incorrectly assumes tension must exceed weight for support - equal forces produce zero net force and zero acceleration, maintaining rest. For hanging mass problems in equilibrium, tension equals weight.

Question 4

1. Static friction acts downward because it always opposes the applied push.
2. The normal force equals $mg$ because the block is at rest.
3. Static friction must balance $mg$ in the vertical direction. (correct answer)
4. The applied push equals $mg$ because horizontal and vertical forces must match in magnitude.

Explanation: 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.

Question 5

1. Static friction points up the incline, opposing the tendency of the block to slide down. (correct answer)
2. Static friction points down the incline, because friction always acts in the direction of gravity.
3. Static friction is zero because the block is not moving.
4. Static friction points perpendicular to the incline, opposing the normal force.

Explanation: 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.

Question 6

1. $15\,\text{N}$ forward (correct answer)
2. $65\,\text{N}$ forward
3. $15\,\text{N}$ backward
4. $0\,\text{N}$ because vertical forces cancel

Explanation: 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.

Question 7

1. The normal force equals the weight because the crate is on a horizontal surface.
2. The friction force equals $T\sin 30^\circ$ because friction balances the vertical component of tension.
3. The friction force equals the horizontal component of the tension, $T\cos 30^\circ$. (correct answer)
4. The net force must point in the direction of motion even at constant velocity.

Explanation: The skill being tested is applying Newton's laws to analyze force balances in systems with inclined tension and friction. Newton's first law states that an object in uniform motion continues unless acted on by a net force, the second law equates net force to mass times acceleration, and the third describes action-reaction pairs. In this system, the crate moves at constant velocity, so the net force is zero in both horizontal and vertical directions per Newton's first law. The friction force equals the horizontal component of tension, T cos 30°, because it balances the forward pull to maintain zero net horizontal force, making choice C correct. Choice D is incorrect because at constant velocity, acceleration is zero, so net force must be zero, not in the direction of motion. For similar questions, draw a free-body diagram and resolve all forces into horizontal and vertical components. Ensure the sum of forces in each direction equals zero if there's no acceleration.

Question 8

1. The block must accelerate because two forces are acting on it.
2. The net horizontal force is zero, consistent with no acceleration. (correct answer)
3. The weight cancels one of the tensions, producing equilibrium.
4. The block is in equilibrium only if the tensions are greater than the weight.

Explanation: The skill being tested is recognizing equilibrium with balanced opposing forces. Newton's first law states zero net force for no motion change, the second equates force to m a, and the third describes reactions. The block is at rest with equal opposing tensions, so net horizontal force is zero. This consistency with no acceleration supports choice B. Choice A is incorrect because balanced forces can result in zero acceleration despite multiple forces. In similar cases, calculate net force by vector sum. Verify equilibrium if observed motion is absent.

Question 9

1. The net force parallel to the incline is positive because the block is moving downhill.
2. The downhill component of gravity equals the kinetic friction magnitude. (correct answer)
3. The normal force equals the component of gravity parallel to the incline.
4. Kinetic friction points downhill because it opposes gravity.

Explanation: The skill being tested is resolving forces on inclines for constant speed motion. Newton's first law implies zero net force for uniform motion, the second relates force to acceleration, and the third involves pairs. The block slides at constant speed, so net force parallel to the incline is zero. The downhill gravity component equals kinetic friction, balancing for no acceleration, supporting choice B. Choice A is incorrect because net parallel force is zero, not positive. In similar incline problems, resolve gravity into parallel and perpendicular components. Ensure friction opposes relative motion and balances for constant speed.

Question 10

1. Static friction acts downward because the hand pushes into the wall.
2. Static friction acts upward and its magnitude equals the book's weight. (correct answer)
3. The normal force from the wall must equal the book's weight.
4. The hand force must be equal to $mg$ to keep the book from falling.

Explanation: The skill being tested is analyzing vertical static friction in equilibrium against a wall. Newton's first law demands zero net force for rest, the second relates force to acceleration, and the third involves reaction pairs. The book is at rest, so net vertical and horizontal forces are zero. Static friction acts upward equaling the weight to balance gravity, as horizontal forces cancel, supporting choice B. Choice C is incorrect because the normal force balances the hand force horizontally, not the weight vertically. In like questions, separate forces into components perpendicular to each other. Check equilibrium by ensuring opposing forces sum to zero in each direction.

Question 11

1. Static friction is $18\,\text{N}$ because that is the maximum possible.
2. Static friction is $12\,\text{N}$ opposing the applied force. (correct answer)
3. Static friction is zero because the box is not moving.
4. The net force is $12\,\text{N}$ to the right, but acceleration is prevented by inertia.

Explanation: 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.

Question 12

1. Tension equals weight because the mass is attached to a string.
2. Weight is less than tension because the mass accelerates downward.
3. Weight is greater than tension, producing a downward net force. (correct answer)
4. Tension is zero because the pulley is frictionless.

Explanation: The skill being tested is force analysis in accelerating Atwood systems. Newton's first law for equilibrium, second for net force causing acceleration, third for pairs. The hanging mass accelerates downward, so net force is downward with mg > T. This produces the downward acceleration, fitting choice C. Choice A is incorrect as tension does not equal weight during acceleration. For pulley systems, write equations for each mass. Check consistency by ensuring heavier mass has mg > T if descending.

Question 13

1. Net force is upward because the ball is moving upward.
2. Net force is zero because no contact forces act.
3. Net force is downward with magnitude $mg$. (correct answer)
4. Net force decreases as the ball rises because its speed decreases.

Explanation: 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.

Question 14

1. Each string tension equals $mg$ because both support the block.
2. The vertical components of the two tensions sum to $mg$. (correct answer)
3. The horizontal components of the two tensions sum to $mg$.
4. The net force is upward because tensions act upward.

Explanation: The skill being tested is resolving tensions in symmetric suspension systems. Newton's first law demands zero net force for equilibrium, the second relates to acceleration, and the third pairs. The block is at rest, so vertical force components balance the weight. Vertical components of tensions sum to mg, making choice B correct. Choice A is incorrect as each tension is less than mg due to angles. In similar setups, resolve tensions into vertical and horizontal. Confirm horizontal components cancel in symmetry.

Question 15

1. The wall exerts no force unless the wall moves.
2. The wall's force on the hand is $150\,\text{N}$ opposite the push direction. (correct answer)
3. The wall's force on the hand is greater than $150\,\text{N}$ because the wall is rigid.
4. The wall's force on the hand is an action–reaction pair with the force of gravity on the hand.

Explanation: This question tests Newton's Third Law and force analysis on the hand. Newton's Third Law states that forces between interacting objects are equal in magnitude and opposite in direction. When the hand pushes the wall with 150 N, the wall pushes back on the hand with 150 N in the opposite direction. For the stationary hand, these forces balance. The correct answer applies Newton's Third Law directly. Answer C incorrectly claims the wall exerts more force due to rigidity - Newton's Third Law requires exactly equal magnitudes regardless of object properties. For action-reaction pairs, forces are always equal in magnitude and opposite in direction.

Question 16

1. The upward wrist contact force equals the weight of the held mass. (correct answer)
2. The upward wrist contact force is zero because muscles provide the force internally.
3. The net force is downward because gravity acts and the system is not supported.
4. The wrist contact force must be less than the weight because the hand is not accelerating.

Explanation: 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.

Question 17

1. $N = mg$ because the suitcase is accelerating horizontally.
2. $N > mg$ because the pull adds downward force.
3. $N < mg$ because the pull has an upward component. (correct answer)
4. The normal force is zero because the handle supports the suitcase.

Explanation: This question tests vertical force balance with angled applied forces. When a force is applied at an angle, it has both horizontal and vertical components. The 50 N force at 45° has an upward component of 50 sin 45° ≈ 35.4 N. For no vertical acceleration, vertical forces must balance: N + 50 sin 45° = mg, giving N = mg - 50 sin 45° < mg. The correct answer recognizes that the upward pull component reduces the normal force. Answer A incorrectly ignores the vertical component of the angled force. For angled force problems, always resolve forces into components and apply equilibrium to each direction.

Question 18

1. Static friction is $12\,\text{N}$ to the left, so the net horizontal force is zero. (correct answer)
2. Static friction is zero because the block is not moving.
3. Kinetic friction is $12\,\text{N}$ to the left because friction always opposes applied force.
4. The net horizontal force is $12\,\text{N}$ to the right, but acceleration is prevented by inertia.

Explanation: This question tests understanding of static friction and equilibrium conditions. Newton's First Law requires that for an object at rest, all forces must balance to produce zero net force. Since the block remains stationary despite the 12 N applied force, static friction must provide an equal and opposite 12 N force to the left, maintaining equilibrium. The correct answer recognizes that static friction adjusts to match applied forces up to its maximum value. Answer B is incorrect because static friction can be non-zero for stationary objects - it prevents motion by balancing applied forces. For static equilibrium problems, identify all forces and ensure they sum to zero.

Question 19

1. The scale reads $mg$ because the student is not moving relative to the elevator.
2. The scale reads less than $mg$ because acceleration is upward.
3. The scale reads greater than $mg$ because $N - mg = ma$ upward. (correct answer)
4. The scale reads zero because the elevator is accelerating.

Explanation: 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.

Question 20

1. $9.0\,\text{N}$ to the right (correct answer)
2. $0\,\text{N}$ because the cart starts from rest
3. $29\,\text{N}$ downward
4. $9.0\,\text{N}$ to the left (reaction force)

Explanation: 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.