AP Physics 1 Unit 2: Forces and Newton's Laws — Free Review Games.

F = ma, so a = F/m = 40/10 = 4 m/s^2.

By Newton's First Law, constant velocity means zero net force (forces are balanced).

Weight = mg = 5 x 9.8 = 49 N.

A free-body diagram shows all forces acting on a single isolated object.

The Third Law pair is: Earth pulls the book down (weight), and the book pulls Earth up. The normal force from the table is a separate interaction.

F_parallel = mg*sin(30) = 5(10)(0.5) = 25 N.

Friction = mu_k x Normal force = 0.3 x 100 = 30 N.

a = (m2 - m1)*g / (m1 + m2) = (5-3)(10)/(5+3) = 20/8 = 2.5 m/s^2.

Mass = 2000/10 = 200 kg. Friction = 0.2(2000) = 400 N. Net force = 500 - 400 = 100 N. a = 100/200 = 0.5 m/s^2.

When accelerating upward, the floor must push up harder than gravity (N > mg) to produce the upward acceleration, making you feel heavier.

Max friction on top block = mu_s * m1 * g = 0.4(2)(10) = 8 N. Max acceleration of top block = 8/2 = 4 m/s^2. Max F for system = (m1+m2)(4) = 5(4) = 20 N.

Using equilibrium: T1*cos30 = T2*cos60, and T1*sin30 + T2*sin60 = 40. Solving gives T2 (at 60 degrees) = 20 N.

At minimum speed at the top, tension = 0 so \(mg = mv^2/r\). \(v = \sqrt{gr} = \sqrt{10 \times 2} = \sqrt{20} = 4.47\) m/s.

a = g*sin(37) = 10(0.6) = 6.0 m/s^2.

System acceleration = 12/(1+2+3) = 2 m/s^2. Tension between 1 kg and 2 kg accelerates only the 1 kg block: T = 1(2) = 2 N.

Newton's First Law, the law of inertia, states that motion changes only when a nonzero net external force acts on an object. 'Its own mass' is wrong because mass alone does not cause a change in velocity; it only determines resistance to such change. Students should remember that constant velocity, including zero velocity, always implies zero net force.

The normal force arises from direct physical contact between two surfaces pushing against each other, making it a contact force. Gravitational force between planets acts at a distance without contact, so 'Gravitational force between planets' is incorrect. Recognizing which forces require contact versus act at a distance helps correctly draw free-body diagrams.

Newton's Second Law states that the net force on an object equals its mass times its acceleration, expressed as \(F = ma\). The expression \(F = mv\) incorrectly relates force to momentum-like quantity without accounting for acceleration, which is not the correct form of the law. Every dynamics problem in AP Physics 1 begins with correctly identifying net force through $F_{net} = ma$.

Static friction is the force that resists the initiation of relative motion between two surfaces in contact and adjusts up to a maximum value. Kinetic friction, in contrast, applies only once the surfaces are already sliding relative to each other, so 'Kinetic friction' does not fit this description. Students should remember that static friction can vary while kinetic friction is generally constant for given surfaces.

Newton's Third Law states that for every action force there is a simultaneous, equal-magnitude, and oppositely-directed reaction force, so object B exerts a force on A of equal size but opposite direction. The choice 'Exerts a smaller force on A' is wrong because action-reaction pairs are always equal in magnitude, never smaller or larger. This principle should be applied carefully, remembering that the two forces act on different objects and never cancel each other.

In free-body diagrams, longer arrows represent forces of greater magnitude, giving a visual sense of relative force sizes. The option 'The direction of the force only' is incomplete because direction is shown by the arrow's orientation, not its length. Drawing arrows proportionally helps students visually check whether forces are balanced or unbalanced.

The newton (N) is the SI unit of force, defined as the force needed to accelerate a 1 kg mass at \(1\,\text{m/s}^2\). A joule measures energy, not force, so 'Joule' is incorrect in this context. Recognizing correct units helps students verify that calculated answers are dimensionally consistent on the AP exam.

Because the car moves at constant velocity, Newton's First Law requires that the net force be zero, even though individual forces like engine thrust and friction still act on the car. The idea that the net force is 'Equal to the car's weight' is wrong because weight is a single vertical force, not the net of all forces including horizontal ones. This reinforces that constant velocity, not zero velocity, still implies equilibrium.

Mass is the measure of an object's inertia, meaning greater mass means greater resistance to acceleration for a given force, consistent with \(F=ma\). Weight depends on gravity and location, so it is not the intrinsic property that resists acceleration, making 'Weight' incorrect. Students should distinguish mass, an unchanging scalar property, from weight, which is a force that varies with gravitational field.

For a stationary box on a horizontal surface, the downward weight is balanced by the upward normal force, resulting in zero vertical acceleration. Friction acts horizontally in this scenario, so pairing 'Weight and friction' does not correctly represent the vertical balance. Recognizing which forces act along which axis is essential to constructing accurate free-body diagrams.

Pushing down at an angle adds a downward component of force that the surface must counteract, so the normal force increases beyond just the block's weight to maintain equilibrium. The claim that the normal force 'stays the same' ignores the added vertical downward component from the applied push. Students should always sum vertical force components, not just weight, to determine the true normal force.

Since the two forces are perpendicular, the net force magnitude is found using the Pythagorean theorem: \(\sqrt{6^2+8^2}=\sqrt{100}=10\,\text{N}\). Simply adding the values to get 'a value of \(14\,\text{N}\)' is incorrect because forces at right angles do not add linearly like scalars. This illustrates the importance of vector addition when combining forces that are not collinear.

Using Newton's Second Law, \(a=\frac{F}{m}=\frac{18}{6}=3\,\text{m/s}^2\), which correctly relates net force, mass, and acceleration. The answer '\(0.33\,\text{m/s}^2\)' incorrectly inverts the formula by dividing mass by force instead of force by mass. Students should always confirm which quantity is being solved for before rearranging \(F=ma\).

The net horizontal force is \(15-3=12\,\text{N}\), so acceleration equals \(\frac{12}{3}=4\,\text{m/s}^2\)... wait check: 12/3=4, so correct answer should be 4. Let me redo.

The net force is the tension minus friction, \(15-3=12\,\text{N}\), and dividing by mass gives \(a=\frac{12}{3}=4\,\text{m/s}^2\). The choice '\(5\,\text{m/s}^2\)' would result from forgetting to subtract friction correctly, showing why every opposing force must be included in the net force calculation. This problem emphasizes that net force, not applied force alone, determines acceleration.

The component of gravity parallel to the incline surface is \(mg\sin\theta\), since this is the component that causes the block to slide along the slope. The expression \(mg\cos\theta\) actually represents the component perpendicular to the incline, which balances the normal force instead. Breaking gravity into parallel and perpendicular components relative to the incline is a key technique for incline problems.

At constant velocity the applied force equals friction, so \(\mu_k=\frac{F}{N}=\frac{50}{10\times9.8}\approx0.51\). The choice '\(0.10\)' would result from ignoring gravitational acceleration and using only mass in the denominator, which is dimensionally incorrect. This shows that friction coefficients are dimensionless ratios of force, always calculated using the normal force, not mass alone.

The eastward and westward forces cancel exactly since they are equal and opposite, leaving only the \(5\,\text{N}\) north force as the net force. The answer '\(0\,\text{N}\)' incorrectly assumes all forces cancel, ignoring the unopposed northward component. Students should analyze each direction independently before combining components into a final net force vector.

Applying Newton's Second Law vertically, \(T-mg=ma\), so \(T=m(g+a)=2(9.8+2)=23.6\,\text{N}\). The value '\(19.6\,\text{N}\)' is simply the object's weight and ignores the additional tension needed to accelerate it upward. This problem shows that apparent weight increases when acceleration is directed upward, a common elevator scenario on the AP exam.

For most surfaces, the coefficient of static friction is greater than or equal to the coefficient of kinetic friction, because more force is typically needed to start motion than to sustain it. The claim that '\(\mu_k\) is always greater than \(\mu_s\)' contradicts the physical behavior observed in nearly all real materials. Students should remember this relationship when determining whether an object will begin to move under an applied force.

The horizontal component of the applied force is \(F\cos\theta = 20\cos30^\circ \approx 17.3\,\text{N}\), using standard vector decomposition. The value '\(10\,\text{N}\)' actually corresponds to the vertical component, \(F\sin30^\circ\), showing a common mix-up between sine and cosine components. Careful labeling of angle reference lines prevents this common error on force decomposition problems.

Since the string only needs to accelerate the trailing block, and both blocks share the same acceleration with equal mass, the tension works out to be half the total applied force needed to accelerate both masses together. Claiming 'Tension equals the full applied force' ignores that the leading block also requires part of that force to accelerate itself. This kind of two-block system reinforces treating connected objects both as a system and individually to solve for internal forces like tension.

The acceleration is \(a=\frac{\Delta v}{\Delta t}=\frac{20}{5}=4\,\text{m/s}^2\), so the net force is \(F=ma=1000\times4=4000\,\text{N}\). The choice '\(200\,\text{N}\)' incorrectly divides mass by time rather than properly computing acceleration first. This problem shows the importance of finding acceleration from kinematics data before applying Newton's Second Law.

Just before an object begins to slide, static friction reaches its maximum value, \(f_{s,max}=\mu_sN\), balancing the gravitational component along the incline. Saying the force 'is zero' is incorrect because friction must actively be resisting the tendency to slide at this critical point. This threshold condition is often used to determine the coefficient of static friction experimentally.

A rocket's forward acceleration results directly from the reaction force to the gas it expels downward, a clear demonstration of equal and opposite action-reaction forces. A ball rolling to a stop primarily illustrates friction opposing motion, not an action-reaction pair between two distinct interacting objects. Recognizing action-reaction pairs, which always act on different objects, is essential for correctly applying Newton's Third Law.

At constant velocity, whether stationary or moving, the net force is zero, so tension in the scale must still equal the object's weight of \(50\,\text{N}\). The answer 'Less than \(50\,\text{N}\)' incorrectly assumes downward motion alone reduces tension, but only acceleration, not velocity, changes tension. This reinforces that constant velocity always implies equilibrium regardless of direction or speed.

Since \(N=mg\cos\theta\), solving gives \(\cos\theta = \frac{400}{500}=0.8\), so \(\theta=\cos^{-1}(0.8)\approx36.9^\circ\). The value '\(53.1^\circ\)' is the complementary angle, which would result from confusing sine and cosine relationships on an incline. This underscores the need to correctly identify which trigonometric function relates the normal force to the incline angle.

A common and serious free-body diagram error is forgetting to include gravity, which acts on virtually every object near Earth's surface unless explicitly stated otherwise. Including only forces acting on the object of interest is actually correct practice, not an error, since a free-body diagram should isolate one object at a time. Students should systematically check for gravity, normal force, friction, tension, and applied forces every time they construct a diagram.

Treating the system together, \(a=\frac{m_2g}{m_1+m_2}=\frac{3\times9.8}{5}=5.88\)... let me recompute: 3*9.8=29.4, /5=5.88. So answer should be 5.88, matching choice 0 not 1. Need to fix.

Treating both masses as one system, the net driving force is the hanging weight \(m_2g=29.4\,\text{N}\) divided by total mass \(m_1+m_2=5\,\text{kg}\), giving \(a=5.88\,\text{m/s}^2\). The choice '\(9.8\,\text{m/s}^2\)' would be correct only for free fall, but here the hanging mass is restrained by the connected block, reducing the acceleration. This pulley-system approach of combining masses is a standard technique for connected-object problems.

The system accelerates at \(a=\frac{12}{6}=2\,\text{m/s}^2\), and analyzing the 4 kg block alone, the contact force must supply \(F=ma=4\times2=8\,\text{N}\). The answer '\(12\,\text{N}\)' incorrectly assumes the entire applied force is transmitted through contact, ignoring that some force accelerates the pushing block itself. This two-block contact-force method is frequently tested to check understanding of internal forces within a connected system.

Since the applied force has an upward vertical component \(F\sin\theta\), the normal force decreases to maintain vertical equilibrium, giving \(N=mg-F\sin\theta\). The simple expression '\(N=mg\)' ignores the vertical component of the angled applied force, which is a common oversight in incline-free horizontal problems with angled forces. This demonstrates that normal force is not always simply equal to weight whenever other forces have vertical components.

The maximum acceleration before slipping is limited by friction on the top block, \(a_{max}=\mu g=0.4\times9.8=3.92\,\text{m/s}^2\), and applying this to the total mass gives \(F=(16)(3.92)\approx62.7\,\text{N}\). The value '\(23.5\,\text{N}\)' would result from applying friction force to only the top block's mass rather than the combined system needed to reach the shared maximum acceleration. This layered-block problem requires linking the friction limit on one block to the acceleration of the entire system.

When a pulley has rotational inertia, the two tensions must differ to produce the net torque required to angularly accelerate the pulley, meaning tension is larger on the side that must do more work turning the pulley, typically the hanging-mass side. The claim that 'Tension is equal on both sides regardless of the pulley's inertia' only holds for idealized massless, frictionless pulleys, not this more realistic scenario. Advanced pulley systems remind students that idealized assumptions like massless pulleys are simplifications not always valid in more complex analyses.

At constant velocity, the friction force exactly balances the gravitational component along the incline, so \(\mu_k mg\cos\theta = mg\sin\theta\), which simplifies to \(\mu_k=\tan\theta\). The expression '\(\mu_k = \sin\theta\)' incorrectly omits dividing by the cosine term needed to isolate the friction coefficient from the normal force relation. This derivation is a classic technique for experimentally measuring \(\mu_k\) using only the critical sliding angle.

Decelerating while moving downward means the elevator's acceleration points upward, so by Newton's Second Law the normal force, and thus the scale reading, must exceed the person's true weight to produce this upward acceleration. The answer 'Scale reads less than true weight' would apply if the elevator were accelerating downward, not decelerating from downward motion. This distinction between velocity direction and acceleration direction is critical for correctly solving elevator-related apparent weight problems.

The net driving force without friction is \((m_2-m_1)g=2\times9.8=19.6\,\text{N}\), and subtracting the opposing \(5\,\text{N}\) friction-like force gives \(14.6\,\text{N}\), so \(a=\frac{14.6}{10}=1.46\,\text{m/s}^2\). The value '\(1.96\,\text{m/s}^2\)' is the standard Atwood machine result ignoring the additional opposing force, which is incorrect once axle friction is introduced. This problem shows how extra resistive forces must be incorporated into the net force even in idealized systems like Atwood machines.

If an external applied force has a component directed up the slope, it can supplement friction to keep the block in equilibrium even beyond the usual critical angle determined by \(\mu_s\) alone. The claim 'Reduced block mass' is incorrect because mass cancels out of the critical angle condition \(\tan\theta=\mu_s\), so changing mass alone would not change whether the block stays still. This question emphasizes that equilibrium conditions must include all applied forces, not just gravity, normal force, and friction, when solving incline problems.

For circular motion on a flat curve, static friction provides centripetal force, so \(\mu_s mg = \frac{mv^2}{r}\), giving \(v=\sqrt{\mu_s g r}=\sqrt{0.6\times9.8\times50}\approx17.1\,\text{m/s}\). The value '\(29.4\,\text{m/s}\)' incorrectly omits the square root step, treating the product directly as the speed rather than as \(v^2\). This problem connects Newton's Second Law with circular motion, a synthesis frequently tested on the AP Physics 1 exam.

Along the incline, \(F - mg\sin37^\circ = ma\), so \(F = m(a+g\sin37^\circ)=10(2+5.88)=78.8\,\text{N}\). The value '\(20\,\text{N}\)' comes from using only \(ma\) and forgetting to add the gravitational component along the incline that must also be overcome. Multi-step incline problems require combining the net force equation with the gravitational component along the slope before solving for the unknown applied force.

When a car accelerates, static friction between the tires and road actually pushes the car forward, meaning friction here provides the driving force rather than resisting motion. In the case of 'A book sliding to a stop on a table', friction opposes the book's existing motion instead of causing acceleration in the direction of travel. This example teaches that friction's role depends on context, sometimes enabling motion rather than only resisting it.

The horizontal acceleration relates to the rope angle by \(a=g\tan\theta=9.8\times\tan10^\circ\approx1.73\,\text{m/s}^2\), derived from balancing horizontal and vertical tension components. The value '\(0.98\,\text{m/s}^2\)' would come from mistakenly using \(g\sin\theta\) instead of \(g\tan\theta\), missing the correct trigonometric relationship between the two force components. This pendulum-in-accelerating-frame setup is a classic technique for indirectly measuring acceleration using an inclined hanging mass.

For the block to remain stationary relative to the frictionless wedge surface, the horizontal acceleration of the system must satisfy \(a=g\tan\theta\), balancing the normal force components exactly against gravity in the block's reference frame. The claim that 'The applied force must equal \(Mg\)' is unrelated to the geometric condition needed to prevent relative sliding on a frictionless incline. This classic problem combines Newton's Second Law with non-inertial reasoning, a synthesis often required for the most challenging AP Physics 1 questions.

At the point just before sliding, the applied force equals the maximum static friction, so \(\mu_s = \frac{f_{s,max}}{N}=\frac{9.8}{2\times9.8}=0.5\). The value '\(1.0\)' would result from forgetting to divide by the normal force entirely and instead treating the applied force as directly equal to the coefficient. This experimental method of finding the threshold sliding force is a standard way to measure \(\mu_s\) in a physics laboratory setting.

A movable pulley splits the supporting force across two string segments, effectively halving the tension needed in each segment to support the same hanging weight compared to a single fixed pulley system. The claim 'The tension is the same as in the standard setup' ignores the mechanical advantage introduced by the movable pulley, which fundamentally changes the force distribution. This concept of mechanical advantage through pulley systems extends basic Newton's law analysis into more advanced simple-machine reasoning valuable for synthesis-level AP questions.