Forces and Newton's Laws — Free Physics Review Games.

Newton's First Law (law of inertia) states that an object maintains its state of motion unless acted upon by a net external force.

The Newton (N) is the SI unit of force, defined as kg*m/s^2.

Inertia is the tendency of an object to resist changes in its velocity; more massive objects have greater inertia.

Newton's Second Law states that net force equals mass times acceleration: F = ma.

Friction is a contact force that opposes the relative motion or tendency of motion between two surfaces.

A free-body diagram isolates one object and shows all the forces (as arrows) acting on it to help analyze motion.

Using F = ma: a = F/m = 50/10 = 5 m/s^2.

Newton's Third Law states that when object A exerts a force on object B, B exerts an equal and opposite force on A.

The normal force is the contact force exerted by a surface perpendicular to that surface, supporting the object against gravity.

Static friction keeps objects from starting to move and is typically greater than kinetic friction, which acts on objects already in motion.

Net force = 30 - 10 = 20 N. Acceleration = 20/5 = 4 m/s^2.

While gravitational force increases with mass (F = mg), inertia also increases proportionally, keeping acceleration constant at g.

Apparent weight = m(g + a) = 60(9.8 + 2) = 60(11.8) = 708 N.

Tension is the pulling force transmitted along a rope, string, or cable when forces are applied at its ends.

Total acceleration = 16/(3+5) = 2 m/s^2. Tension on 3 kg block: T = 3 x 2 = 6 N.

Newton's First Law states that an object's velocity changes only when a net, unbalanced force acts on it, so an object at rest stays at rest until such a force is applied. Gravity alone is wrong because gravity can be balanced by another force such as the normal force, producing no net force and no change in motion. This law establishes the concept of inertia, which is central to understanding when and why objects change their state of motion.

Newton's Third Law states that for every action force there is an equal and opposite reaction force, which is exactly what happens when a swimmer pushes water backward and is propelled forward by the water's reaction force. The choice describing a book resting on a table illustrates force balance and Newton's First Law, not an action-reaction pair between two different objects. Remembering that action-reaction pairs always act on two different objects, never the same one, helps distinguish this law from equilibrium situations.

A free-body diagram uses arrows to represent every force acting on an isolated object, with arrow length indicating relative magnitude and direction indicating the force's direction. The path of travel is not shown in a free-body diagram because the diagram isolates forces, not motion history. Correctly identifying all force vectors is the first step in applying Newton's Second Law to solve for acceleration.

Kinetic friction is the resistive force that acts between two surfaces already in relative motion, opposing the direction of sliding. Static friction is incorrect because it only applies before motion begins, preventing an object from starting to slide. Recognizing whether an object is moving or stationary tells you which friction coefficient and force equation to apply.

In the equation \(f = \mu N\), \(\mu\) is the coefficient of friction, a dimensionless number describing how rough or smooth the interaction is between two surfaces. Mass is not represented by \(\mu\) since mass appears separately in Newton's Second Law, not in the friction formula itself. Students should remember that \(\mu\) depends only on the surface materials in contact, not on the object's speed or size.

Weight is the gravitational force on an object, calculated as \(W = mg\), which follows directly from Newton's Second Law with acceleration equal to \(g\). The option \(W = \frac{m}{g}\) is incorrect because dividing mass by acceleration would give units inconsistent with force. Students should always distinguish mass, a measure of matter that stays constant, from weight, a force that depends on the local gravitational field.

Inertia is the tendency of any object to resist a change in its velocity, whether that means starting to move, stopping, or changing direction, and it is directly related to an object's mass. The force pulling objects toward Earth is gravity, a distinct concept from inertia even though both influence motion. Understanding inertia helps explain why more massive objects require greater force to achieve the same acceleration as lighter ones.

The normal force is a contact force exerted by a surface perpendicular to that surface, preventing objects from passing through it. It does not always point downward since its direction depends on the surface's orientation, such as pointing upward on flat ground or sideways against a vertical wall. Recognizing that the normal force adjusts to balance other perpendicular forces is essential for correctly drawing free-body diagrams.

Newton's First Law explains that an object with zero net force remains in its current state of motion, and since the normal force from the table balances gravity, the box stays at rest. Newton's Second Law is not the best explanation here because it describes how forces cause acceleration, but in this case there is no acceleration to explain. Recognizing balanced forces as evidence of zero net force is a key skill for analyzing equilibrium situations.

A true action-reaction pair consists of two equal and opposite forces acting on two different objects simultaneously, exactly like the hammer striking the nail and the nail pushing back on the hammer. The pairing of the book's weight with the table's normal force is incorrect as a Third Law pair because both forces act on the same object, the book, making them a balanced-force pair, not an action-reaction pair. A useful test is to check whether the two forces act on different objects and arise from the same type of interaction.

Using Newton's Second Law, \(F = ma = 4\, \text{kg} \times 3\, \text{m/s}^2 = 12\, \text{N}\), so the net force acting on the object is 12 newtons. The value \(1.33\, \text{N}\) comes from mistakenly dividing mass by acceleration instead of multiplying, which produces the wrong units and result. Always confirm that force calculations multiply mass and acceleration together according to \(F=ma\).

On a horizontal surface, the normal force equals the object's weight, so \(f = \mu N = 0.3 \times 50\, \text{N} = 15\, \text{N}\). The answer \(150\, \text{N}\) is incorrect because it results from dividing instead of multiplying, or mixing up the coefficient with a whole-number factor. Students should remember that on flat, horizontal surfaces, normal force equals weight only when there is no vertical acceleration or additional vertical force.

The component of gravity parallel to an incline is \(mg\sin\theta = 2 \times 10 \times \sin(30^\circ) = 20 \times 0.5 = 10\, \text{N}\). The value \(17.3\, \text{N}\) is incorrect because it uses \(\cos(30^\circ)\) instead of \(\sin(30^\circ)\), which actually gives the perpendicular component of gravity, not the parallel one. Breaking gravity into components along and perpendicular to an incline is a critical skill for solving inclined-plane problems.

The net force is \(8\, \text{N} - 3\, \text{N} = 5\, \text{N}\) to the right, so acceleration equals \(a = F/m = 5/2.5 = 2\, \text{m/s}^2\) to the right, the same direction as the net force. The option \(11\, \text{m/s}^2\) is wrong because it incorrectly adds the two forces instead of subtracting the opposing one. When forces act in opposite directions, always subtract to find the net force before applying \(F=ma\).

Static friction increases to match an applied force up to a maximum value, and the crate begins sliding exactly when the applied force exceeds this maximum static friction force. A ball rolling down a frictionless ramp is incorrect because there is no friction at all in that scenario, so static friction cannot be at play. This principle explains why it often takes more force to start moving an object than to keep it moving once it is already sliding.

A correct free-body diagram of a book on an incline includes gravity pulling straight down, the normal force perpendicular to the ramp's surface, and friction acting parallel to the surface if the surfaces are not frictionless. Including only gravity ignores the contact forces from the ramp that are essential to correctly analyze the book's motion or equilibrium. Every contact and non-contact force acting on the object must be represented to correctly apply Newton's Second Law.

By Newton's Third Law, the rocket pushes the exhaust gases backward with a certain force, and the gases simultaneously push the rocket forward with an equal and opposite force, propelling it. The idea that gases 'pull' the rocket through friction is incorrect because there is no medium for friction in the vacuum of space, and the rocket still accelerates forward. Recognizing that thrust arises from momentum exchange, not from pushing against air, is essential for correctly explaining rocket propulsion.

At constant velocity, the applied force equals the friction force, so \(f = 18\, \text{N}\), and since \(N = mg = 60\, \text{N}\), the coefficient is \(\mu = f/N = 18/60 = 0.3\). The value \(3.0\) is incorrect because it results from dividing normal force by friction force rather than friction by normal force. Constant velocity always signals zero net force, meaning applied force and friction are exactly balanced.

Static friction coefficients are typically higher than kinetic friction coefficients for the same pair of surfaces, meaning more force is required to overcome static friction and start motion than to sustain it once moving. The claim that kinetic friction always exceeds static friction is factually backward and misrepresents how friction transitions from static to kinetic. This distinction is a common exam topic, so students should remember \(\mu_s > \mu_k\) generally holds true.

Equilibrium means the net force is zero, so the vector sum of all three forces, considering both magnitude and direction, must equal zero even if the individual forces have different magnitudes and directions. Requiring all magnitudes to be equal is incorrect because equilibrium depends on vector addition, not on the forces having identical sizes. This concept is foundational for solving multi-force free-body diagram problems using components.

Since the elevator is stationary, the net force on the mass is zero, so tension equals weight: \(T = mg = 3 \times 9.8 = 29.4\, \text{N}\). The reading of \(3\, \text{N}\) is incorrect because it mistakenly reports the mass value in newtons rather than calculating the actual gravitational force. Whenever an object hangs motionless, tension in the supporting string always equals the object's weight.

Newton's Second Law applies to the net force, which is the vector sum of every individual force acting on the object, and this net force determines acceleration through $F_{net} = ma$. The claim that only the largest force determines acceleration is wrong because smaller opposing or perpendicular forces still contribute to the vector sum and change the resulting acceleration. Students must always combine all forces vectorially before applying \(F=ma\) to avoid calculation errors.

According to Newton's First Law, the crate's inertia keeps it at its original velocity until an external force, such as friction from the truck bed, acts on it, so as the truck accelerates forward the crate appears to lag behind and slide backward relative to the truck. There is no mysterious backward force acting on the crate from the ground observer's perspective, only the crate's tendency to resist a change in motion. This illustrates why sudden accelerations can cause unsecured objects to seem to move opposite to the vehicle's acceleration.

Using Newton's Second Law, \(a = F/m = 4000/1000 = 4\, \text{m/s}^2\), representing the magnitude of the car's deceleration. The value \(0.25\, \text{m/s}^2\) is incorrect because it comes from inverting the formula and dividing mass by force instead of force by mass. Always ensure units and formula structure, \(a = F/m\), are applied consistently when solving for acceleration.

When an elevator accelerates upward, the net force must also point upward, so the normal force from the floor must exceed the person's weight to produce that upward acceleration, following \(N - mg = ma\). The claim that the normal force equals the person's weight exactly is only true when the elevator moves at constant velocity or is stationary, not while accelerating upward. This concept explains the sensation of feeling heavier when an elevator first accelerates upward.

The tension in the connecting rope equals the force required to accelerate only the back crate at the system's shared acceleration, since the rope is the sole horizontal force acting on the rear crate: $T = m_{back} a$. The tension is not equal to the total applied force on the entire system because part of that force is used to accelerate the front crate as well. Analyzing connected objects requires isolating each mass and applying Newton's Second Law separately to find internal forces like tension.

Once the ball leaves the thrower's hand and air resistance is ignored, the only force acting on it is gravity, so the free-body diagram shows a single downward arrow representing the ball's weight. The idea of a forward 'throw force' persisting is a common misconception, since no continuing force is needed to sustain horizontal motion once the ball leaves the hand, according to Newton's First Law. This example emphasizes that objects in projectile motion experience only gravity as a net force when air resistance is negligible.

For the block to remain motionless, the gravitational component pulling it down the incline, \(mg\sin\theta\), must be less than or equal to the maximum static friction force, \(\mu mg\cos\theta\), which resists sliding. The inequality \(mg\cos\theta \leq \mu mg\sin\theta\) incorrectly swaps the roles of the parallel and perpendicular gravity components, reversing the physical meaning of the condition. This inequality is fundamental for determining the critical angle at which objects begin to slide on inclined surfaces.

Since the push has an upward vertical component of \(40\sin(37^\circ) = 24\, \text{N}\), the normal force is reduced from the weight: \(N = mg - 40\sin(37^\circ) = 50 - 24 = 26\, \text{N}\). The answer \(74\, \text{N}\) is incorrect because it wrongly adds the vertical component of the push instead of subtracting it, which would only happen if the push were directed downward. When an applied force has an upward component, students must subtract that component from weight to find the correct reduced normal force.

For this Atwood-style system, acceleration is $a = \frac{m_{hanging} g}{m_{total}} = \frac{4 \times 10}{10} = 4\, \text{m/s}^2$, since the hanging mass's weight is the only net driving force on the combined system of 10 kg. The value \(10\, \text{m/s}^2\) is incorrect because it mistakenly uses gravitational acceleration directly rather than dividing the hanging weight by the total system mass. When analyzing connected systems, always divide the net driving force by the total mass being accelerated, not just one component's mass.

Newton's First Law explains that the passenger's body continues moving forward at its original velocity due to inertia, since no external force acts on the passenger until the dashboard applies a stopping force. Newton's Second Law is not the best explanation, because it describes how forces cause acceleration but does not primarily explain why the passenger keeps moving forward in the absence of a restraining force. This scenario is the classic real-world demonstration of inertia and the importance of seatbelts in providing the needed external force to decelerate passengers safely.

The maximum friction force on the top block is $\mu m_{top} g = 0.4 \times 10 \times 10 = 40\, \text{N}$, which limits its acceleration to \(a = 40/10 = 4\, \text{m/s}^2\); applying this acceleration to the total mass gives $F = (m_{top}+m_{bottom})a = 30 \times 4 = 120\, \text{N}$. The value \(40\, \text{N}\) is incorrect because it only accounts for the force needed to accelerate the top block, ignoring that the same acceleration must also be produced in the bottom block. This type of stacked-block problem requires finding the maximum shared acceleration limited by friction, then applying that acceleration to the entire system's mass.

Along the incline, \(T - mg\sin\theta = ma\), so \(T = m(a + g\sin\theta) = 12(1.5 + 9.8 \times 0.342) \approx 12(1.5+3.35) \approx 58.2\, \text{N}\). The value \(18\, \text{N}\) is incorrect because it only accounts for \(ma\) and ignores the gravitational component pulling the sled back down the slope. On inclines, tension must overcome both the component of gravity along the slope and provide the additional force needed for acceleration.

To solve for tension in two ropes at different angles, each tension must be broken into horizontal and vertical components, then the vertical components must sum to equal the object's weight while horizontal components cancel to zero, since the object is in equilibrium. Dividing the weight equally between the ropes is only valid when the ropes are symmetric with equal angles, which is not the case here since the angles differ. This component-based method is the standard technique for solving any multi-rope or multi-force equilibrium problem where angles are unequal.

The maximum static friction force is \(\mu_s mg = 0.5 \times 3 \times 10 = 15\, \text{N}\), which exceeds the applied \(12\, \text{N}\), so static friction can fully balance the applied force and the block remains stationary. The answer \(1\, \text{m/s}^2\) is incorrect because it assumes the block is already moving and mistakenly applies kinetic friction, which does not apply until motion actually begins. Students must always compare the applied force to maximum static friction first before assuming an object starts moving.

Setting maximum static friction equal to the required centripetal force, \(f = \frac{mv^2}{r}\), gives \(9000 = \frac{1500 v^2}{50}\), so \(v^2 = 300\) and \(v \approx 17.3\, \text{m/s}\). The value \(600\, \text{m/s}\) is incorrect because it comes from failing to take the square root after solving for \(v^2\). Although circular motion connects to friction concepts, students must remember that friction here provides the centripetal force that keeps the car on its curved path.

The normal force from the wall equals the horizontal applied force, \(60\, \text{N}\), so maximum static friction is \(\mu N = 0.4 \times 60 = 24\, \text{N}\), which is greater than the block's \(20\, \text{N}\) weight, meaning the block does not slide down. The claim that friction cannot act vertically is incorrect because friction always acts parallel to the contact surface, and in this case the wall's vertical surface allows friction to act upward, opposing gravity. This problem shows that friction can act in any direction parallel to a surface, not just horizontally on the ground.

The net driving force is the hanging weight minus friction on the table block: $F_{net} = m_2 g - \mu m_3 g = (2)(10) - 0.2(3)(10) = 20 - 6 = 14\, \text{N}$, and dividing by total mass gives \(a = 14/5 = 2.8\, \text{m/s}^2\). The value \(4\, \text{m/s}^2\) is incorrect because it ignores the friction force acting against the motion of the block on the table. Solving connected-system problems with friction requires including all resistive forces before dividing by the total mass.

After the push, no external force acts on the astronaut in the vacuum of deep space, so according to Newton's First Law, the astronaut continues moving at constant velocity indefinitely. The idea that the spacecraft continues exerting force is wrong, because the push was a single momentary interaction, and forces do not persist after contact ends. This scenario illustrates Newton's First Law in an idealized, friction-free, gravity-free environment, which is useful for understanding inertia without the interference of everyday resistive forces.

Newton's Third Law guarantees that the box pushes back on the person with exactly \(50\, \text{N}\) regardless of whether the box moves, but this action-reaction pair does not by itself explain the box's stillness, which is instead explained by the box's static friction with the floor balancing the applied force (Newton's First Law). The claim that the box exerts less force back because it is heavier misunderstands Newton's Third Law, since reaction forces are always equal in magnitude regardless of mass. Students often confuse Third Law force pairs with the separate concept of force balance, so it's important to treat these as two different explanations.

For a frictionless banked curve, \(\tan\theta = \frac{v^2}{rg} = \frac{225}{400} = 0.5625\), giving \(\theta \approx 29.5^\circ\). The answer \(45^\circ\) is incorrect because it would require \(\tan\theta = 1\), corresponding to a much higher speed or smaller radius than given. This banked-curve relationship shows how the horizontal component of the normal force alone can supply the centripetal force needed for circular motion.

Using energy conservation, the kinetic energy at the bottom equals $mgh$, and this energy is dissipated by friction over distance \(d\): $mgh = \mu_k mg d$, which simplifies to \(d = \frac{h}{\mu_k}\). The answer \(\mu_k h\) is incorrect because it inverts the relationship, implying a larger friction coefficient produces a longer stopping distance, which contradicts the physical reality that more friction stops objects sooner. Combining energy methods with Newton's Laws and friction concepts is a common technique for solving multi-stage motion problems.

The hanging block's weight (\(2 \times 10 = 20\, \text{N}\)) pulls the system toward the pulley while the \(10\, \text{N}\) force pulls the other way, giving a net force of \(20 - 10 = 10\, \text{N}\) toward the pulley, but since the question defines the applied force direction as positive, the acceleration is \(-10/6 \approx -1.67\, \text{m/s}^2\), meaning \(1.67\, \text{m/s}^2\) toward the pulley (opposite the applied force direction as stated). The answer \(5\, \text{m/s}^2\) ignores the \(10\, \text{N}\) opposing force entirely and only accounts for the hanging weight. This problem demonstrates that all forces on a connected system, including any externally applied forces, must be combined before dividing by total mass.

For equilibrium, the vector sum of all three forces must be zero, and since \(F_1\) and \(F_2\) are equal in magnitude and separated by \(120^\circ\), their resultant is \(10\, \text{N}\) at \(60^\circ\); the third force must be equal in magnitude but opposite, placing it at \(60^\circ + 180^\circ = 240^\circ\). The option \(0\, \text{N}\) is incorrect because the first two forces do not already cancel each other out, so a nonzero third force is required to achieve equilibrium. Balancing three or more non-collinear forces requires careful vector addition using components or symmetry arguments.

At constant velocity, the net force is zero, so tension must exactly balance the gravitational component along the slope: \(T = mg\sin\theta = 50 \times 9.8 \times \sin(25^\circ) \approx 50 \times 9.8 \times 0.423 \approx 207\, \text{N}\). The value \(490\, \text{N}\) is incorrect because it represents the skier's full weight without resolving it into the component actually acting along the slope's direction. Constant velocity on an incline always implies that the pulling force exactly cancels the parallel component of gravity, not the total weight.