Newton's Second Law (F=ma)

28 soru

Soru 1Soru →

A military recovery vehicle is using a winch to pull a disabled transport truck out of deep mud on a level surface. The disabled truck has a weight of 50,000 N50,000 \text{ N}. The mud exerts a constant resistive drag force of 12,000 N12,000 \text{ N} against the truck's motion. If the winch must accelerate the truck horizontally at a rate of 0.5 m/s20.5 \text{ m/s}^2, how much tension force must the winch cable provide? (Assume gravitational acceleration g=10 m/s2g = 10 \text{ m/s}^2)

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Cevap: 14,500 N14,500 \text{ N}

Cevap

The winch cable must provide 14,500 N14,500 \text{ N} of tension force.
The correct answer is found by first converting the truck's weight (50,000 N50,000 \text{ N}) to its mass (5,000 kg5,000 \text{ kg}) using g=10 m/s2g = 10 \text{ m/s}^2. Next, the net force required to accelerate this mass at 0.5 m/s20.5 \text{ m/s}^2 is calculated as 2,500 N2,500 \text{ N} using F=maF = ma. Finally, because the mud exerts a 12,000 N12,000 \text{ N} opposing drag, the winch must pull with enough force to overcome the drag and still provide the 2,500 N2,500 \text{ N} net force. Therefore, the total required tension is 12,000 N+2,500 N=14,500 N12,000 \text{ N} + 2,500 \text{ N} = 14,500 \text{ N}.

Adım Adım Çözüm

1
Determine the mass of the disabled truck from its given weight.
m=Wg=50,000 N10 m/s2=5,000 kgm = \frac{W}{g} = \frac{50,000 \text{ N}}{10 \text{ m/s}^2} = 5,000 \text{ kg}
Newton's second law (F=maF = ma) requires mass, not weight.
2
Calculate the required net force to achieve the target acceleration.
Fnet=ma=5,000 kg×0.5 m/s2=2,500 NF_{\text{net}} = ma = 5,000 \text{ kg} \times 0.5 \text{ m/s}^2 = 2,500 \text{ N}
This is the net force that must remain after all opposing forces are overcome.
3
Set up the force balance equation for the horizontal direction.
Fnet=Twinch−FdragF_{\text{net}} = T_{\text{winch}} - F_{\text{drag}}
The net force is the vector sum of the applied tension pulling forward and the drag force resisting backward.
4
Solve the equation for the required winch tension (TwinchT_{\text{winch}}).
2,500 N=Twinch−12,000 N  ⟹  Twinch=14,500 N2,500 \text{ N} = T_{\text{winch}} - 12,000 \text{ N} \implies T_{\text{winch}} = 14,500 \text{ N}
Adding the drag force to the required net force gives the total applied force.

Anahtar Kavram

Newton's Second Law applied to multi-force systems.
Soru 2Soru →

Match each mechanical operational scenario to its resulting acceleration using Newton's Second Law (F=maF = ma).

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Öğeler

A 1,200 kg1,200 \text{ kg} aerospace capsule is tested in a track, propelled by a net horizontal force of 3,600 N3,600 \text{ N}.
A 1,500 kg1,500 \text{ kg} forklift experiences a net force of 2,250 N2,250 \text{ N} while accelerating on a warehouse floor.
A 2,500 kg2,500 \text{ kg} bridge section is positioned by cranes applying a net lateral force of 1,250 N1,250 \text{ N}.
A 4,000 kg4,000 \text{ kg} submarine ballast drops through the ocean, experiencing a net downward force of 16,000 N16,000 \text{ N}.

Eşleşmeler

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Cevap

The correct pairings are determined by dividing the net force by the mass for each scenario: the aerospace capsule accelerates at 3.0 m/s23.0 \text{ m/s}^2, the forklift accelerates at 1.5 m/s21.5 \text{ m/s}^2, the bridge section accelerates at 0.5 m/s20.5 \text{ m/s}^2, and the submarine ballast accelerates at 4.0 m/s24.0 \text{ m/s}^2.
Newton's Second Law dictates that acceleration is directly proportional to net force and inversely proportional to mass (a=F/ma = F/m). By dividing the net force by the mass in each of the four scenarios, the correct acceleration values are directly calculated and paired.

Adım Adım Çözüm

1
Identify the known variables provided in each scenario.
Each scenario provides the net force (FF) in Newtons and the object's mass (mm) in kilograms.
Newton's Second Law requires both net force and mass to determine an object's resulting acceleration.
2
Rearrange Newton's Second Law (F=maF = ma) to solve for the missing variable.
The formula becomes a=F/ma = F/m.
Because the right items represent acceleration, the equation must be isolated for acceleration (aa) before computing.
3
Calculate the specific acceleration for each mass and force pairing.
Capsule: 3,600/1,200=3.0 m/s23,600 / 1,200 = 3.0 \text{ m/s}^2. Forklift: 2,250/1,500=1.5 m/s22,250 / 1,500 = 1.5 \text{ m/s}^2. Bridge section: 1,250/2,500=0.5 m/s21,250 / 2,500 = 0.5 \text{ m/s}^2. Ballast: 16,000/4,000=4.0 m/s216,000 / 4,000 = 4.0 \text{ m/s}^2.
Dividing the specific force by the given mass maps each scenario to exactly one value in the right column.

Anahtar Kavram

Solving for acceleration using Newton's Second Law (a=F/ma = F/m)
Soru 3Soru →

An autonomous warehouse transport vehicle with a mass of 2,000 kg2,000 \text{ kg} experiences a constant frictional resistance of 2,000 N2,000 \text{ N} while moving across a level concrete floor. When its electric motors produce a forward thrust of 6,000 N6,000 \text{ N}, the vehicle accelerates at a steady rate. If the motors are commanded to double their forward thrust to 12,000 N12,000 \text{ N} while the frictional resistance remains unchanged, what is the vehicle's new acceleration?

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Cevap: 5.0 m/s25.0 \text{ m/s}^2

Cevap

The new acceleration is 5.0 m/s25.0 \text{ m/s}^2.
According to Newton's Second Law, acceleration is determined by the net force acting on an object (a=Fnetma = \frac{F_{net}}{m}). The net force is the forward thrust minus the friction. With a new thrust of 12,000 N12,000 \text{ N} and a constant friction of 2,000 N2,000 \text{ N}, the net force is 10,000 N10,000 \text{ N}. Dividing this by the mass of 2,000 kg2,000 \text{ kg} yields an acceleration of 5.0 m/s25.0 \text{ m/s}^2.

Adım Adım Çözüm

1
Identify the forces acting on the vehicle in the new scenario.
Forward thrust (FthrustF_{thrust}) = 12,000 N12,000 \text{ N}. Frictional resistance (FfrictionF_{friction}) = 2,000 N2,000 \text{ N}.
Newton's Second Law requires the net force, which is the sum of all opposing forces.
2
Calculate the new net force.
Fnet=12,000 N−2,000 N=10,000 NF_{net} = 12,000 \text{ N} - 2,000 \text{ N} = 10,000 \text{ N}.
The net force is the actual force available to accelerate the mass.
3
Apply Newton's Second Law (F=maF = ma) to find the new acceleration.
a=Fnetm=10,000 N2,000 kg=5.0 m/s2a = \frac{F_{net}}{m} = \frac{10,000 \text{ N}}{2,000 \text{ kg}} = 5.0 \text{ m/s}^2.
Acceleration is equal to the net force divided by the mass.

Anahtar Kavram

Acceleration is directly proportional to the net force, not just the applied force. To find acceleration, all opposing forces must be subtracted from the applied force before dividing by mass.
Soru 4Soru →

A twin-engine military cargo aircraft with a total mass of 80,000 kg80,000 \text{ kg} is accelerating down a runway for takeoff. Each of the two engines generates 140,000 N140,000 \text{ N} of forward thrust. If the combined forces of rolling friction and aerodynamic drag opposing the aircraft's motion amount to 80,000 N80,000 \text{ N}, what is the forward acceleration of the aircraft in m/s2\text{m/s}^2?

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Cevap: 2.5

Cevap

The forward acceleration of the aircraft is 2.5 m/s².
To accurately determine the acceleration, one must first find the net force acting on the aircraft. The total forward thrust is the sum of both engines (140,000 N×2=280,000 N140,000 \text{ N} \times 2 = 280,000 \text{ N}). Subtracting the opposing forces of friction and drag (80,000 N80,000 \text{ N}) yields a net forward force of 200,000 N200,000 \text{ N}. Dividing this net force by the aircraft's mass of 80,000 kg80,000 \text{ kg} gives a forward acceleration of 2.5 m/s22.5 \text{ m/s}^2.

Adım Adım Çözüm

1
Calculate the total forward thrust applied to the aircraft.
2×140,000 N=280,000 N2 \times 140,000 \text{ N} = 280,000 \text{ N}
The aircraft has two engines, and the problem provides the thrust generated by each individual engine.
2
Calculate the net force (FnetF_{net}) acting on the aircraft.
Fnet=280,000 N−80,000 N=200,000 NF_{net} = 280,000 \text{ N} - 80,000 \text{ N} = 200,000 \text{ N}
Net force is the vector sum of all forces. The opposing forces (drag and friction) must be subtracted from the forward thrust.
3
Apply Newton's Second Law to calculate the acceleration.
a=200,000 N80,000 kg=2.5 m/s2a = \frac{200,000 \text{ N}}{80,000 \text{ kg}} = 2.5 \text{ m/s}^2
According to Newton's Second Law, acceleration equals the net force divided by the mass (a=Fnetma = \frac{F_{net}}{m}).

Anahtar Kavram

Newton's Second Law of Motion (Fnet=maF_{net} = ma)
Soru 5Soru →

An arresting gear system is used to stop a 12,000 kg12,000 \text{ kg} aircraft landing on a flight deck. The aircraft touches down with a velocity of 144 km/h144 \text{ km/h} and is brought to a complete stop in 2.0 seconds2.0 \text{ seconds} by the arresting wire. Assuming a constant deceleration, what is the magnitude of the average net braking force applied to the aircraft?

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Cevap: 240,000 N240,000 \text{ N}

Cevap

The correct magnitude of the net braking force is 240,000 N240,000 \text{ N}.
The correct answer accurately calculates the force by first converting the velocity from km/h to m/s (144 km/h×1000 m3600 s=40 m/s144 \text{ km/h} \times \frac{1000 \text{ m}}{3600 \text{ s}} = 40 \text{ m/s}). It then finds the magnitude of acceleration (a=Δvt=40 m/s2.0 s=20 m/s2a = \frac{\Delta v}{t} = \frac{40 \text{ m/s}}{2.0 \text{ s}} = 20 \text{ m/s}^2) and applies Newton's Second Law (F=m×a=12,000 kg×20 m/s2=240,000 NF = m \times a = 12,000 \text{ kg} \times 20 \text{ m/s}^2 = 240,000 \text{ N}).

Adım Adım Çözüm

1
Convert the initial velocity from km/h to m/s.
144 km/h×(1000 m3600 s)=40 m/s144 \text{ km/h} \times \left(\frac{1000 \text{ m}}{3600 \text{ s}}\right) = 40 \text{ m/s}
Standard SI units (meters and seconds) are required to correctly calculate force in Newtons.
2
Calculate the magnitude of acceleration (deceleration) using the change in velocity over time.
a=40 m/s2.0 s=20 m/s2a = \frac{40 \text{ m/s}}{2.0 \text{ s}} = 20 \text{ m/s}^2
Acceleration is defined as the rate of change of velocity.
3
Apply Newton's Second Law to find the average net braking force.
F=12,000 kg×20 m/s2=240,000 NF = 12,000 \text{ kg} \times 20 \text{ m/s}^2 = 240,000 \text{ N}
Newton's Second Law states that force equals mass times acceleration (F=maF = ma).

Anahtar Kavram

Newton's Second Law (F=ma) and Kinematics
Soru 6Soru →

A naval torpedo with a mass of 1,200 kg1,200 \text{ kg} is fired from a submarine. If the propulsion system generates a constant net forward force of 6,000 N6,000 \text{ N}, what is the forward acceleration of the torpedo?

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Cevap: 5.0 m/s25.0 \text{ m/s}^2

Cevap

The correct answer is 5.0 m/s25.0 \text{ m/s}^2.
According to Newton's Second Law (F=maF = ma), acceleration is calculated by dividing the net force by the mass. Dividing 6,000 N6,000 \text{ N} by 1,200 kg1,200 \text{ kg} yields 5.0 m/s25.0 \text{ m/s}^2.

Adım Adım Çözüm

1
Identify the given values from the problem statement.
Mass (mm) = 1,200 kg1,200 \text{ kg}, Net Force (FF) = 6,000 N6,000 \text{ N}.
Extracting the known variables is necessary to apply the correct physics formula.
2
Apply Newton's Second Law of Motion.
The formula is F=maF = ma. Rearranging to solve for acceleration gives a=Fma = \frac{F}{m}.
Newton's Second Law relates force, mass, and acceleration.
3
Substitute the values into the formula and solve.
a=60001200=5.0 m/s2a = \frac{6000}{1200} = 5.0 \text{ m/s}^2.
Dividing the net force by the mass computes the resulting acceleration.

Anahtar Kavram

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F=maF = ma).
Soru 7Soru →

Match each physical scenario with the magnitude of its resulting acceleration. Apply Newton's Second Law of Motion (F=maF=ma) and ensure you calculate the net force for each situation.

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Öğeler

A 4,000 kg transport vehicle on a level surface experiences a forward thrust of 12,000 N, a headwind drag of 2,000 N, and rolling friction of 2,000 N.
A 1,500 kg hovercraft on water is pushed by two rear propellers providing 3,000 N of thrust each, while facing a 1,500 N opposing water current.
A 500 kg payload is hoisted vertically upward by a crane whose cable exerts an upward tension of 8,000 N. (Assume g=10 m/s2g = 10 \text{ m/s}^2)
An 800 kg elevator descends as its support cable applies an upward tension of 4,800 N. (Assume g=10 m/s2g = 10 \text{ m/s}^2)

Eşleşmeler

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Cevap

The correct matches are: Transport vehicle maps to 2.0 m/s², Hovercraft maps to 3.0 m/s², Hoisted payload maps to 6.0 m/s², and Descending elevator maps to 4.0 m/s².
Each correct match is derived by first identifying all acting forces on the object, determining the net force (FnetF_{net}), and then dividing by the mass (a=Fnet/ma = F_{net}/m). For horizontal motion, opposing forces (friction, drag) are subtracted from applied thrust. For vertical motion, the object's weight (mgmg) must be calculated and subtracted from the cable tension.

Adım Adım Çözüm

1
Calculate the acceleration for the transport vehicle.
Net force = 12,000 N - 2,000 N - 2,000 N = 8,000 N. Acceleration = 8,000 N / 4,000 kg = 2.0 m/s².
Forces acting in the opposite direction of motion (drag and friction) must be subtracted from the forward thrust.
2
Calculate the acceleration for the hovercraft.
Net force = (2 * 3,000 N) - 1,500 N = 4,500 N. Acceleration = 4,500 N / 1,500 kg = 3.0 m/s².
The total thrust from both propellers must be combined before subtracting the opposing water current.
3
Calculate the acceleration for the hoisted payload.
Weight = 500 kg * 10 m/s² = 5,000 N. Net upward force = 8,000 N - 5,000 N = 3,000 N. Acceleration = 3,000 N / 500 kg = 6.0 m/s².
In vertical motion, the downward force of gravity (weight) must be calculated and subtracted from the upward tension to find the net force.
4
Calculate the acceleration for the descending elevator.
Weight = 800 kg * 10 m/s² = 8,000 N. Net downward force = 8,000 N - 4,800 N = 3,200 N. Acceleration = 3,200 N / 800 kg = 4.0 m/s².
Because the elevator descends, the downward gravitational force exceeds the upward tension. Net force is weight minus tension.

Anahtar Kavram

Newton's Second Law of Motion (Fnet=maF_{net} = ma)
Soru 8Soru →

Using Newton's Second Law (F=maF = ma), match each mechanical scenario with its correct missing value (force, mass, or acceleration).

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Öğeler

Net force required to accelerate a 15 kg15 \text{ kg} tool box at 3 m/s23 \text{ m/s}^2
Mass of a component that accelerates at 4 m/s24 \text{ m/s}^2 when a 150 N150 \text{ N} push is opposed by 30 N30 \text{ N} of friction
Acceleration of a 250 kg250 \text{ kg} engine block pushed with 1,200 N1,200 \text{ N} of forward force against 200 N200 \text{ N} of friction
The new acceleration of a moving object if the net force acting on it is tripled (assuming mass is constant) from an initial state that produced 2 m/s22 \text{ m/s}^2

Eşleşmeler

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Cevap

The matches are calculated by rearranging F=maF = ma to solve for the missing variable, factoring in net force where applicable, and applying direct proportionality.
Newton's Second Law states that net force equals mass times acceleration (F=maF = ma). By determining the net force and algebraically rearranging the formula, or by applying proportional reasoning rules, each scenario can be correctly matched to its missing value.

Adım Adım Çözüm

1
Calculate the net force where opposing forces are present.
Subtract friction from the applied force to find the true net force acting on the object.
Newton's Second Law only applies to the net force, not just the applied force.
2
Apply Newton's Second Law (F=maF = ma) or its rearranged forms.
Use F=maF = ma for force, m=Fam = \frac{F}{a} for mass, and a=Fma = \frac{F}{m} for acceleration.
These equations isolate the missing variable for the first three scenarios.
3
Apply proportional reasoning for the final scenario.
Recognize that tripling the force triples the acceleration.
Since a=Fma = \frac{F}{m}, acceleration and force are directly proportional when mass is constant.

Anahtar Kavram

Newton's Second Law
Tahmini Süre:1m 30s
Soru 9Soru →

A military transport truck with a mass of 8,000 kg8,000 \text{ kg} is towing a loaded artillery trailer with a mass of 4,000 kg4,000 \text{ kg} along a level, straight road. The truck's engine generates a total forward driving force of 36,000 N36,000 \text{ N}. During motion, the truck experiences a resistive drag force of 8,000 N8,000 \text{ N}, and the trailer experiences its own resistive drag force of 4,000 N4,000 \text{ N}. Assuming the tow bar connecting them is perfectly rigid and horizontal, what is the tension force in the tow bar?

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Cevap: 12,000 N12,000 \text{ N}

Cevap

The tension force in the tow bar is 12,000 N12,000 \text{ N}.
To find the tension, you must analyze the forces. First, calculate the acceleration of the entire system. The total mass is 12,000 kg12,000 \text{ kg}. The net force on the system is the driving force minus both drag forces: 36,000 N−(8,000 N+4,000 N)=24,000 N36,000 \text{ N} - (8,000 \text{ N} + 4,000 \text{ N}) = 24,000 \text{ N}. Using a=Fnet/ma = F_{net}/m, the system's acceleration is 24,000 N/12,000 kg=2.0 m/s224,000 \text{ N} / 12,000 \text{ kg} = 2.0 \text{ m/s}^2. Next, isolate the trailer to find the tension. The net force required to accelerate just the trailer is 4,000 kg×2.0 m/s2=8,000 N4,000 \text{ kg} \times 2.0 \text{ m/s}^2 = 8,000 \text{ N}. The tension must provide this net force PLUS overcome the trailer's 4,000 N4,000 \text{ N} drag. Therefore, Tension = Net Force + Drag = 8,000 N+4,000 N=12,000 N8,000 \text{ N} + 4,000 \text{ N} = 12,000 \text{ N}.

Adım Adım Çözüm

1
Calculate the total mass and the net forward force for the entire two-vehicle system.
Total mass = 12,000 kg12,000 \text{ kg}. Net force = 36,000 N−(8,000 N+4,000 N)=24,000 N36,000 \text{ N} - (8,000 \text{ N} + 4,000 \text{ N}) = 24,000 \text{ N}.
Newton's Second Law must first be applied to the system as a whole to determine their shared acceleration.
2
Calculate the acceleration of the system using Newton's Second Law (a=Fnet/ma = F_{net}/m).
a=24,000 N/12,000 kg=2.0 m/s2a = 24,000 \text{ N} / 12,000 \text{ kg} = 2.0 \text{ m/s}^2.
Because the tow bar is rigid, the truck and trailer accelerate at the exact same rate.
3
Isolate the trailer as a free body and apply Newton's Second Law to find the tension (TT) in the tow bar.
The net force on the trailer is m×a=4,000 kg×2.0 m/s2=8,000 Nm \times a = 4,000 \text{ kg} \times 2.0 \text{ m/s}^2 = 8,000 \text{ N}. Since Fnet=T−FdragF_{net} = T - F_{drag}, we get 8,000 N=T−4,000 N8,000 \text{ N} = T - 4,000 \text{ N}, which means T=12,000 NT = 12,000 \text{ N}.
The tension in the tow bar must provide enough force to both overcome the trailer's drag and accelerate its mass.

Anahtar Kavram

Newton's Second Law applied to Multi-Body Systems
Soru 10Soru →

During a manufacturing process, a pneumatic pusher applies a constant net horizontal force of 300 N300 \text{ N} to slide a steel component across a frictionless track. If the component accelerates at 15 m/s215 \text{ m/s}^2, what is the mass of the steel component?

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Cevap: 20 kg20 \text{ kg}

Cevap

20 kg20 \text{ kg}
According to Newton's Second Law of Motion, the net force on an object is equal to its mass times its acceleration (F=maF = ma). To find the mass when both force and acceleration are known, the formula must be rearranged to divide the net force by the acceleration (m=Fam = \frac{F}{a}). Dividing the given 300 N300 \text{ N} by 15 m/s215 \text{ m/s}^2 yields a mass of 20 kg20 \text{ kg}.

Adım Adım Çözüm

1
Identify the given variables provided in the scenario.
Force (FF) = 300 N300 \text{ N}, Acceleration (aa) = 15 m/s215 \text{ m/s}^2.
Establishing the known physical quantities is required before selecting and applying the correct formula.
2
State Newton's Second Law of Motion and rearrange it algebraically to solve for mass.
The formula F=maF = ma becomes m=Fam = \frac{F}{a}.
The question asks for the mass of the component, so the standard equation must be isolated for the variable mm.
3
Substitute the known values into the rearranged equation and calculate the final result.
m=30015=20 kgm = \frac{300}{15} = 20 \text{ kg}.
Provides the final mass of the steel component based on the applied force and resulting acceleration.

Anahtar Kavram

Newton's Second Law (F=maF=ma)
Soru 11Soru →

A compact utility tractor applies a constant net forward force of 1,200 N1,200 \text{ N} to push a stalled passenger vehicle on a level, paved road. If the vehicle accelerates at a rate of 1.5 m/s21.5 \text{ m/s}^2, what is the mass of the vehicle?

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Cevap: 800

Cevap

800 kg
According to Newton's Second Law (F=maF=ma), the mass of an object is equal to the net force acting upon it divided by its acceleration. By dividing the applied net force of 1,200 N1,200 \text{ N} by the acceleration of 1.5 m/s21.5 \text{ m/s}^2, we determine the mass of the vehicle is 800 kg800 \text{ kg}.

Adım Adım Çözüm

1
Identify the known variables provided in the problem.
Net Force (FF) = 1,200 N1,200 \text{ N}, Acceleration (aa) = 1.5 m/s21.5 \text{ m/s}^2.
Extracting the given information is the first step to determine which physics formula to apply.
2
State the formula for Newton's Second Law of Motion.
F=maF = ma
This formula establishes the mathematical relationship between net force, mass, and acceleration.
3
Rearrange the formula to solve for the unknown variable, mass (mm).
m=Fam = \frac{F}{a}
To find the mass, it must be isolated algebraically on one side of the equation.
4
Substitute the known values into the rearranged formula and compute the result.
m=1,2001.5=800m = \frac{1,200}{1.5} = 800
Executing the division calculation provides the final mass in kilograms.

Anahtar Kavram

Newton's Second Law of Motion (F=ma)
Soru 12Soru →

An aviation mechanic pushes a loaded tool cart across a level hangar floor. If the mechanic applies a net forward force of 120 N120 \text{ N} and the cart accelerates at 2 m/s22 \text{ m/s}^2, what is the mass of the cart?

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Cevap: 60 kg60 \text{ kg}

Cevap

The mass of the tool cart is 60 kg60 \text{ kg}.
According to Newton's Second Law (F=maF = ma), mass is equal to net force divided by acceleration. Dividing 120 N120 \text{ N} by 2 m/s22 \text{ m/s}^2 results in 60 kg60 \text{ kg}.

Adım Adım Çözüm

1
Identify the given variables from the scenario.
Net force (FF) = 120 N120 \text{ N}, Acceleration (aa) = 2 m/s22 \text{ m/s}^2.
These are the known values required to solve for the missing variable.
2
Apply Newton's Second Law of Motion.
The formula is F=maF = ma.
This law defines the physical relationship between net force, mass, and acceleration.
3
Rearrange the formula to solve for mass (mm).
m=Fam = \frac{F}{a}
To find the unknown mass, it must be isolated on one side of the equation.
4
Substitute the values and calculate the final result.
m=1202=60 kgm = \frac{120}{2} = 60 \text{ kg}
Performing the division yields the actual mass of the object.

Anahtar Kavram

Newton's Second Law states that force equals mass times acceleration (F=maF = ma), meaning an object's mass can be calculated by dividing the net force by its acceleration.
Soru 13Soru →

Match each mechanical scenario to its corresponding missing value (net force, mass, or acceleration) using Newton's Second Law (F=maF = ma).

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

A 10 kg10 \text{ kg} gear assembly accelerating at 3 m/s23 \text{ m/s}^2 requires a net force of...
A mechanism experiencing 100 N100 \text{ N} of net force and accelerating at 5 m/s25 \text{ m/s}^2 has a mass of...
A 25 kg25 \text{ kg} engine block pushed with a net force of 50 N50 \text{ N} experiences an acceleration of...
A 200 kg200 \text{ kg} payload accelerating at 4 m/s24 \text{ m/s}^2 requires a net force of...

Eşleşmeler

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Cevap

Match the scenarios by calculating the missing variable: 10 kg10 \text{ kg} at 3 m/s23 \text{ m/s}^2 matches 30 N30 \text{ N}; 100 N100 \text{ N} at 5 m/s25 \text{ m/s}^2 matches 20 kg20 \text{ kg}; 25 kg25 \text{ kg} at 50 N50 \text{ N} matches 2 m/s22 \text{ m/s}^2; 200 kg200 \text{ kg} at 4 m/s24 \text{ m/s}^2 matches 800 N800 \text{ N}.
The correct matches are found by identifying the two known variables in Newton's Second Law (F=maF = ma) for each scenario and algebraically solving for the third unknown variable.

Adım Adım Çözüm

1
Calculate the net force for the gear assembly.
F=10 kg×3 m/s2=30 NF = 10 \text{ kg} \times 3 \text{ m/s}^2 = 30 \text{ N}
Applying the direct formula F=maF = ma to find the required force.
2
Calculate the mass of the mechanism.
m=100 N5 m/s2=20 kgm = \frac{100 \text{ N}}{5 \text{ m/s}^2} = 20 \text{ kg}
Rearranging the formula to m=Fam = \frac{F}{a} to solve for mass.
3
Calculate the acceleration of the engine block.
a=50 N25 kg=2 m/s2a = \frac{50 \text{ N}}{25 \text{ kg}} = 2 \text{ m/s}^2
Rearranging the formula to a=Fma = \frac{F}{m} to solve for acceleration.
4
Calculate the net force for the payload.
F=200 kg×4 m/s2=800 NF = 200 \text{ kg} \times 4 \text{ m/s}^2 = 800 \text{ N}
Applying the direct formula F=maF = ma to find the required force.

Anahtar Kavram

Newton's Second Law (F=maF = ma)
Soru 14Soru →

A warehouse automated sorting system applies a constant net horizontal force of 200 N200 \text{ N} to a package with a mass of 50 kg50 \text{ kg}. What is the resulting acceleration of the package?

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Cevap: 4 m/s24 \text{ m/s}^2

Cevap

The resulting acceleration of the package is 4 m/s24 \text{ m/s}^2.
According to Newton's Second Law (F=maF = ma), acceleration is calculated by dividing the applied net force by the object's mass. Dividing 200 N200 \text{ N} by 50 kg50 \text{ kg} correctly results in 4 m/s24 \text{ m/s}^2.

Adım Adım Çözüm

1
Identify the known variables provided in the problem.
Net Force (FF) = 200 N200 \text{ N}; Mass (mm) = 50 kg50 \text{ kg}.
Correctly identifying the givens ensures the proper values are placed into the physics formula.
2
State Newton's Second Law of Motion and rearrange it to solve for acceleration.
The formula F=maF = ma becomes a=Fma = \frac{F}{m}.
The question asks for the resulting acceleration, so the formula must isolate aa.
3
Substitute the known values into the rearranged formula and calculate the result.
a=20050=4 m/s2a = \frac{200}{50} = 4 \text{ m/s}^2.
Executing the division yields the final acceleration value in standard metric units.

Anahtar Kavram

Newton's Second Law (F=maF = ma) demonstrates that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Tahmini Süre:30s
Soru 15Soru →

A deck winch on a naval vessel applies a constant net horizontal force of 4,500 N4,500 \text{ N} to drag a heavy supply pallet across a level deck. If the pallet has a mass of 1,500 kg1,500 \text{ kg}, what is the acceleration of the pallet?

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Cevap: 3

Cevap

The correct answer is 3.
By applying Newton's Second Law (F=maF = ma) and rearranging for acceleration (a=Fma = \frac{F}{m}), we divide the net force of 4,500 N4,500 \text{ N} by the mass of 1,500 kg1,500 \text{ kg} to find an acceleration of 3 m/s23 \text{ m/s}^2.

Adım Adım Çözüm

1
Identify the given information
Force (FF) = 4,500 N4,500 \text{ N}, Mass (mm) = 1,500 kg1,500 \text{ kg}
Extracting the known variables is the first step in solving a physics calculation problem.
2
Select the appropriate formula
F=maF = ma
Newton's Second Law relates force, mass, and acceleration.
3
Rearrange the formula to solve for acceleration
a=Fma = \frac{F}{m}
We need to isolate the unknown variable, acceleration (aa).
4
Substitute the values and calculate
a=45001500=3 m/s2a = \frac{4500}{1500} = 3 \text{ m/s}^2
Plugging in the given numbers yields the final numerical answer.

Anahtar Kavram

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass (F=maF = ma).
Soru 16Soru →

A combat engineer squad is using a motorized winch to pull a 2,500 kg2,500 \text{ kg} concrete barrier horizontally across rough ground to construct a defensive position. The winch applies a constant tension of 12,000 N12,000 \text{ N} to the towing cable. If the friction between the barrier and the ground opposes the motion with a force of 7,000 N7,000 \text{ N}, what is the acceleration of the barrier?

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Cevap: 2.0 m/s22.0 \text{ m/s}^2

Cevap

2.0 m/s22.0 \text{ m/s}^2
The correct answer is found by first calculating the net force, which is the applied tension minus the opposing friction (12,000 N−7,000 N=5,000 N12,000 \text{ N} - 7,000 \text{ N} = 5,000 \text{ N}). Then, using Newton's Second Law (a=Fma = \frac{F}{m}), the net force is divided by the mass of the barrier (5,000 N2,500 kg\frac{5,000 \text{ N}}{2,500 \text{ kg}}) to yield an acceleration of 2.0 m/s22.0 \text{ m/s}^2.

Adım Adım Çözüm

1
Identify the applied force and the opposing force acting on the barrier.
Applied Force = 12,000 N12,000 \text{ N}, Opposing Friction = 7,000 N7,000 \text{ N}.
To find the net force, all horizontal forces acting on the object must be accounted for.
2
Calculate the net force (FnetF_{net}) acting on the concrete barrier.
Fnet=12,000 N−7,000 N=5,000 NF_{net} = 12,000 \text{ N} - 7,000 \text{ N} = 5,000 \text{ N}.
Net force is the vector sum of all forces; opposite forces are subtracted.
3
Apply Newton's Second Law (F=maF = ma) to find the acceleration.
a=Fnetm=5,000 N2,500 kg=2.0 m/s2a = \frac{F_{net}}{m} = \frac{5,000 \text{ N}}{2,500 \text{ kg}} = 2.0 \text{ m/s}^2.
Acceleration is directly proportional to net force and inversely proportional to mass.

Anahtar Kavram

Newton's Second Law (F=ma) and Net Force
Soru 17Soru →

An automated cargo tug applies a forward thrust of 24,000 N24,000 \text{ N} to a 1,000 kg1,000 \text{ kg} supply sled. If the ground surface creates a constant opposing frictional force of 4,000 N4,000 \text{ N}, what is the resulting acceleration of the sled?

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Cevap: 20 m/s220 \text{ m/s}^2

Cevap

20 m/s220 \text{ m/s}^2
The sled experiences a forward thrust of 24,000 N24,000 \text{ N} and a backward frictional force of 4,000 N4,000 \text{ N}, resulting in a net force of 20,000 N20,000 \text{ N}. Using Newton's Second Law (F=maF = ma), dividing the net force (20,000 N20,000 \text{ N}) by the mass (1,000 kg1,000 \text{ kg}) gives an acceleration of 20 m/s220 \text{ m/s}^2.

Adım Adım Çözüm

1
Calculate the net force acting on the sled.
Fnet=24,000 N−4,000 N=20,000 NF_{\text{net}} = 24,000 \text{ N} - 4,000 \text{ N} = 20,000 \text{ N}
Acceleration is determined by the net force, which is the applied force minus any opposing forces like friction.
2
Apply Newton's Second Law (F=maF = ma) to find the acceleration.
a=Fnetm=20,000 N1,000 kg=20 m/s2a = \frac{F_{\text{net}}}{m} = \frac{20,000 \text{ N}}{1,000 \text{ kg}} = 20 \text{ m/s}^2
Rearranging the formula to solve for acceleration gives a=Fma = \frac{F}{m}.

Anahtar Kavram

Newton's Second Law (F=maF = ma)
Tahmini Süre:45s
Soru 18Soru →

During a runway clearing operation, an armored recovery vehicle pushes a disabled transport truck horizontally across the tarmac. The truck has a weight of 80,000 N80,000 \text{ N}. The recovery vehicle applies a forward force of 30,000 N30,000 \text{ N}, while the truck's damaged axles create a constant rolling resistance of 10,000 N10,000 \text{ N}. Assuming the acceleration due to gravity is 10 m/s210 \text{ m/s}^2, what is the forward acceleration of the disabled truck?

Cevabı ve açıklamayı göster

Cevap: 2.5 m/s22.5 \text{ m/s}^2

Cevap

The disabled truck will have a forward acceleration of 2.5 m/s22.5 \text{ m/s}^2.
The correct acceleration is found by first converting the truck's weight to its mass, determining the net force by subtracting the rolling resistance from the applied force, and then dividing that net force by the mass.

Adım Adım Çözüm

1
Calculate the mass of the truck from its weight.
m=80,000 N10 m/s2=8,000 kgm = \frac{80,000 \text{ N}}{10 \text{ m/s}^2} = 8,000 \text{ kg}
Newton's Second Law requires mass, not weight, to calculate acceleration.
2
Determine the net force acting on the truck.
Fnet=30,000 N−10,000 N=20,000 NF_{net} = 30,000 \text{ N} - 10,000 \text{ N} = 20,000 \text{ N}
The rolling resistance directly opposes the applied forward force, so it must be subtracted.
3
Calculate the acceleration using Newton's Second Law (F=maF = ma).
a=Fnetm=20,000 N8,000 kg=2.5 m/s2a = \frac{F_{net}}{m} = \frac{20,000 \text{ N}}{8,000 \text{ kg}} = 2.5 \text{ m/s}^2
Acceleration is directly proportional to net force and inversely proportional to mass.

Anahtar Kavram

Applying Newton's Second Law (F=maF=ma) in multi-force scenarios requiring weight-to-mass conversion.
Soru 19Soru →

During a testing sequence, a heavy torpedo with a mass of 1,500 kg1,500 \text{ kg} is fired from a submarine tube. The launch mechanism applies a forward propulsive force of 45,000 N45,000 \text{ N}. If the water resistance creates a constant opposing drag force of 9,000 N9,000 \text{ N}, what is the initial acceleration of the torpedo?

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Cevap: 24

Cevap

24
To find the acceleration, we first determine the net force acting on the torpedo. The forward propulsive force is 45,000 N45,000 \text{ N}, but water resistance pushes back with 9,000 N9,000 \text{ N}. Subtracting the opposing force gives a net force of 36,000 N36,000 \text{ N}. Using Newton's Second Law (F=maF = ma), we rearrange the formula to solve for acceleration (a=Fma = \frac{F}{m}). Dividing the net force (36,000 N36,000 \text{ N}) by the mass (1,500 kg1,500 \text{ kg}) results in an acceleration of 24 m/s224 \text{ m/s}^2.

Adım Adım Çözüm

1
Calculate the net force (FnetF_{net}) acting on the torpedo.
Fnet=45,000 N−9,000 N=36,000 NF_{net} = 45,000 \text{ N} - 9,000 \text{ N} = 36,000 \text{ N}
The torpedo experiences forces in opposite directions, so the drag force must be subtracted from the forward thrust to find the total effective force.
2
Calculate the acceleration (aa) using Newton's Second Law.
a=Fnetm=36,000 N1,500 kg=24 m/s2a = \frac{F_{net}}{m} = \frac{36,000 \text{ N}}{1,500 \text{ kg}} = 24 \text{ m/s}^2
Newton's Second Law states that acceleration equals the net force divided by the mass.

Anahtar Kavram

Newton's Second Law (F=ma) and Net Force
Soru 20Soru →

A military hovercraft with a mass of 6,000 kg6,000 \text{ kg} is transporting equipment across a frozen lake. The hovercraft's thrusters generate a forward force of 27,000 N27,000 \text{ N}, while air resistance and surface friction create a combined opposing force of 9,000 N9,000 \text{ N}. What is the forward acceleration of the hovercraft?

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Cevap: 3.0 m/s23.0 \text{ m/s}^2

Cevap

The correct forward acceleration of the hovercraft is 3.0 m/s23.0 \text{ m/s}^2.
The correct answer accurately determines the net force by subtracting the opposing friction (9,000 N9,000 \text{ N}) from the applied thrust (27,000 N27,000 \text{ N}) to get 18,000 N18,000 \text{ N}. Dividing this net force by the mass (6,000 kg6,000 \text{ kg}) yields an acceleration of 3.0 m/s23.0 \text{ m/s}^2.

Adım Adım Çözüm

1
Calculate the net forward force acting on the hovercraft.
Fnet=27,000 N−9,000 N=18,000 NF_{net} = 27,000 \text{ N} - 9,000 \text{ N} = 18,000 \text{ N}
Acceleration is determined by the net force, which is the sum of all forces acting on the object. The opposing force must be subtracted from the forward thrust.
2
Set up Newton's Second Law equation.
Fnet=m×aF_{net} = m \times a
This formula relates net force, mass, and acceleration.
3
Rearrange the equation to solve for acceleration and substitute the known values.
a=18,000 N6,000 kg=3.0 m/s2a = \frac{18,000 \text{ N}}{6,000 \text{ kg}} = 3.0 \text{ m/s}^2
Dividing the net force by the mass gives the resulting acceleration.

Anahtar Kavram

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass (a=Fnetma = \frac{F_{net}}{m}).
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