Newton’s Laws and Free-Body Diagrams (4A)
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MCAT Chemical and Physical Foundations of Biological Systems › Newton’s Laws and Free-Body Diagrams (4A)
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.
$N<mg$ because the elevator is moving upward.
The net force is zero because only two forces act.
$N>mg$ because the net force must be upward.
$N=mg$ because the person is not moving relative to the elevator.
Explanation
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 $10,\text{kg}$ 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.
$65,\text{N}$ forward
$0,\text{N}$ because vertical forces cancel
$15,\text{N}$ backward
$15,\text{N}$ forward
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.
A $0.50,\text{kg}$ puck slides on ice (negligible friction) with an initial velocity to the right. A constant force of $2.0,\text{N}$ is applied upward (north) for $3,\text{s}$. Based on Newton’s Laws, which statement about the puck’s motion is most consistent?
Forces: applied force (north), weight (down), normal (up).
The puck accelerates east because it already had an eastward velocity.
The net force is zero because vertical forces cancel.
The puck’s speed must remain constant because there is no friction.
The puck accelerates north while continuing to move east, so its velocity direction changes.
Explanation
The skill being tested is applying Newton's laws to changing velocity directions with perpendicular forces. Newton's first law maintains motion without net force, the second defines acceleration from net force, and the third pairs actions. The puck has initial eastward velocity and gains northward acceleration from the applied force, altering its path. The puck accelerates north while moving east, changing velocity direction, which fits choice B. Choice A is incorrect because constant speed does not imply constant velocity if direction changes. For analogous problems, resolve forces perpendicular to initial motion. Confirm by noting acceleration only in the direction of net force.
A $5.0,\text{kg}$ box is pushed across a rough horizontal surface at constant speed. The push is purely horizontal with magnitude $20,\text{N}$. Kinetic friction acts opposite the motion. Which statement is most consistent with Newton’s Laws for the box?
Forces: weight (down), normal (up), applied push (forward), kinetic friction (backward).
Kinetic friction is less than $20,\text{N}$ because the box is moving.
The normal force must be less than the weight because friction is present.
Kinetic friction equals $20,\text{N}$.
The net force equals $20,\text{N}$ in the direction of motion.
Explanation
The skill being tested is understanding force equilibrium at constant speed with friction. Newton's first law requires zero net force for constant velocity, the second links net force to acceleration, and the third covers paired forces. The box moves at constant speed, so net horizontal force is zero, with push balanced by friction. Kinetic friction equals 20 N, opposing the push to maintain equilibrium, making choice B correct. Choice C is incorrect because net force is zero at constant speed, not 20 N forward. For similar scenarios, confirm net force is zero if speed is constant. Analyze by equating opposing forces like applied and friction.
A $12,\text{kg}$ crate rests on a horizontal floor. A student pulls it with a rope at $30^\circ$ above the horizontal with tension $T=60,\text{N}$. The crate moves at constant velocity. Kinetic friction acts between crate and floor. Considering forces on the crate (weight, normal force, tension, friction), which statement about the force balance is most consistent with Newton’s Laws?
(Assume air resistance is negligible.)
The friction force equals the horizontal component of the tension, $T\cos 30^\circ$.
The normal force equals the weight because the crate is on a horizontal surface.
The net force must point in the direction of motion even at constant velocity.
The friction force equals $T\sin 30^\circ$ because friction balances the vertical component of tension.
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.
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).
$0,\text{N}$ because the vertical forces cancel
$14,\text{N}$ to the right
$6,\text{N}$ to the right
$6,\text{N}$ to the left
Explanation
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.
During a biceps curl, the forearm is held stationary at a fixed angle while supporting a weight in the hand. Consider the forearm+hand as the system. External forces include the weight of the dumbbell (down), the weight of the forearm (down), the biceps tendon force (upward component), and the elbow joint contact force. The forearm is not accelerating. Based on Newton’s Laws, which statement is most consistent?
Ignore air resistance; treat the configuration as static.
The vector sum of all external forces on the forearm is zero.
The biceps tendon force must equal the dumbbell weight regardless of geometry.
The net force on the forearm must be nonzero because muscles are pulling.
The elbow joint contact force must be zero because the forearm is stationary.
Explanation
The skill being tested is applying equilibrium conditions to biomechanical systems. Newton's first law requires zero net force for no acceleration, the second defines force-acceleration relation, and the third pairs forces. The forearm is stationary, so vector sum of all external forces is zero. This equilibrium condition makes choice B correct. Choice A is incorrect because muscles provide forces, but net external force is still zero in statics. In similar static systems, sum all external forces to zero. Check torque balance if rotational aspects are involved, but focus on translational here.
A $3.0,\text{kg}$ block on a horizontal surface is attached to a rope. Two students pull on opposite ends of the rope so that the block experiences $15,\text{N}$ tension to the right and $15,\text{N}$ tension to the left (equal magnitudes). The block is observed to remain at rest. Which statement is most consistent with Newton’s Laws?
Forces on block: leftward tension, rightward tension, weight, normal.
The weight cancels one of the tensions, producing equilibrium.
The block must accelerate because two forces are acting on it.
The block is in equilibrium only if the tensions are greater than the weight.
The net horizontal force is zero, consistent with no acceleration.
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.
A $4.0,\text{kg}$ block slides down a rough incline at constant speed. The incline angle is fixed. Forces on the block include gravity, normal force, and kinetic friction. Based on Newton’s Laws, which statement is most consistent with the force components parallel to the incline?
Take the downhill direction along the incline as positive.
The downhill component of gravity equals the kinetic friction magnitude.
Kinetic friction points downhill because it opposes gravity.
The net force parallel to the incline is positive because the block is moving downhill.
The normal force equals the component of gravity parallel to the incline.
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.
A $1.5,\text{kg}$ book is held pressed against a vertical wall by a horizontal hand force $F$ directed into the wall. The book remains at rest. Static friction between the book and wall acts vertically. Which statement about the forces on the book is most consistent with Newton’s Laws?
Forces on the book: weight (down), static friction (up or down), normal force from wall (horizontal), hand force (horizontal).
Static friction acts upward and its magnitude equals the book’s weight.
Static friction acts downward because the hand pushes into the wall.
The hand force must be equal to $mg$ to keep the book from falling.
The normal force from the wall must equal the book’s weight.
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.