Force, Mass, and Motion Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) — Page 2

Question 21Question →

A construction crew uses a pile driver machine that drops a solid steel hammer freely onto a concrete pillar. The operator replaces the original $500\text{-kg}$ hammer with a $1,000\text{-kg}$ hammer and drops it from the exact same height.

Ignoring air resistance and guide rail friction, which of the following statements correctly describes how the final free-fall speed of the heavier hammer compares to the original lighter hammer?

The speeds will be exactly the same, because gravitational acceleration is independent of an object's mass.

The heavier hammer will fall faster, because a larger mass results in a greater gravitational acceleration.

The heavier hammer will fall slower, because its increased mass creates more inertia to resist the downward pull.

The heavier hammer will fall faster, because mass and weight are identical properties that directly dictate an object's falling speed.

Show answer & explanation

Answer: The speeds will be exactly the same, because gravitational acceleration is independent of an object's mass.

Answer

The final speeds will be exactly the same because the acceleration due to gravity is independent of the falling object's mass.

In the absence of air resistance and friction, all objects in free fall experience the exact same uniform acceleration due to gravity, regardless of their mass. Because both hammers are dropped from the same height and accelerate at the same rate, they will achieve the exact same speed just before impact.

Step-by-Step Solution

1

Identify the forces acting on both hammers during their descent.

Since air resistance and friction are ignored, the only force acting on the hammers is the downward pull of gravity.

Establishing that the objects are in a state of free fall is required to determine their rate of acceleration.

2

Relate gravitational force to acceleration using Newton's laws.

The downward force is weight (

F = mg

), and Newton's second law is

F = ma

. Equating them gives

mg = ma

, which simplifies to

a = g

.

This mathematical relationship proves that the mass (

m

) cancels out, meaning acceleration depends only on gravity (

g

).

3

Compare the final velocities of the two hammers.

Both hammers drop from the exact same height with the same constant acceleration. Therefore, their final speeds just before impact must be equal.

According to kinematic equations, identical initial conditions (starting from rest) and identical acceleration over the same distance result in the exact same final velocity.

Key Concept

Independence of Mass in Free Fall

Question 22Question →

A hiker accidentally knocks a rock off the edge of a steep cliff. Assuming there is no air resistance and the acceleration due to gravity is $10\text{ m/s}^2$ , how fast is the rock traveling exactly $3$ seconds after it begins to fall?

Show answer & explanation

Answer: 30

Answer

30

In free fall with no air resistance, an object's speed increases by the acceleration of gravity each second. After

3

seconds of accelerating at

10\text{ m/s}^2

, the speed is exactly

30\text{ m/s}

.

Step-by-Step Solution

1

Identify the relevant formula

v = gt

The rock starts from rest, so its speed depends only on gravitational acceleration multiplied by the elapsed time.

2

Substitute the given values into the formula

v = 10 \times 3 = 30

Gravity accelerates the object by

10\text{ m/s}^2

for every second it falls.

Key Concept

Velocity of an object in free fall increases linearly with time based on gravitational acceleration.

Question 23Question →

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 $50,000 \text{ N}$ . The mud exerts a constant resistive drag force of $12,000 \text{ N}$ against the truck's motion. If the winch must accelerate the truck horizontally at a rate of $0.5 \text{ m/s}^2$ , how much tension force must the winch cable provide? (Assume gravitational acceleration $g = 10 \text{ m/s}^2$ )

22,000 \text{ N}

37,000 \text{ N}

14,500 \text{ N}

2,500 \text{ N}

Show answer & explanation

Answer: $14,500 \text{ N}$

Answer

The winch cable must provide

14,500 \text{ N}

of tension force.

The correct answer is found by first converting the truck's weight (

50,000 \text{ N}

) to its mass (

5,000 \text{ kg}

) using

g = 10 \text{ m/s}^2

. Next, the net force required to accelerate this mass at

0.5 \text{ m/s}^2

is calculated as

2,500 \text{ N}

using

F = ma

. Finally, because the mud exerts a

12,000 \text{ N}

opposing drag, the winch must pull with enough force to overcome the drag and still provide the

2,500 \text{ N}

net force. Therefore, the total required tension is

12,000 \text{ N} + 2,500 \text{ N} = 14,500 \text{ N}

.

Step-by-Step Solution

1

Determine the mass of the disabled truck from its given weight.

m = \frac{W}{g} = \frac{50,000 \text{ N}}{10 \text{ m/s}^2} = 5,000 \text{ kg}

Newton's second law (

F = ma

) requires mass, not weight.

2

Calculate the required net force to achieve the target acceleration.

F_{\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.

F_{\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 (

T_{\text{winch}}

).

2,500 \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.

Key Concept

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

Question 24Question →

Two solid spheres of identical volume and exterior finish—one cast from dense lead and the other from lightweight aluminum—are released simultaneously from a hovering atmospheric research drone. Which of the following best describes their descent characteristics through the air?

The lead sphere will attain a higher terminal velocity because more air resistance is required to counteract its greater weight.

Both spheres will accelerate at an identical rate throughout the entire fall because gravitational acceleration is independent of mass.

The lead sphere will experience a greater rate of acceleration from the exact moment of release because of its larger mass.

The aluminum sphere will take more time to reach its terminal velocity because its lower weight cannot overcome air resistance as quickly.

Show answer & explanation

Answer: The lead sphere will attain a higher terminal velocity because more air resistance is required to counteract its greater weight.

Answer

The lead sphere will attain a higher terminal velocity because more air resistance is required to counteract its greater weight.

An object reaches terminal velocity when the upward force of air resistance exactly equals its downward weight. Since the lead sphere has a much greater mass than the aluminum sphere, it has a proportionally greater weight. Because both spheres have identical dimensions and surface characteristics, they experience the exact same air resistance at any given speed. Therefore, the lead sphere must accelerate to a much higher velocity to generate enough air resistance to balance its larger weight.

Step-by-Step Solution

1

Analyze the forces acting on the spheres at the exact moment of release.

At exactly

t=0

, velocity is zero, so aerodynamic drag is zero. Both spheres accelerate downward at exactly standard gravity (

g

).

Gravity accelerates all objects equally in the absence of air resistance.

2

Determine how air resistance affects each sphere as they gain speed.

Because they are identical in size and shape, both spheres experience the exact same upward drag force at any given descent speed.

Aerodynamic drag depends on velocity, cross-sectional area, and shape, but not on the mass of the object.

3

Identify the conditions required to reach terminal velocity.

Terminal velocity is reached when the upward drag force exactly equals the downward weight.

When the net force is zero, acceleration stops and velocity becomes constant.

4

Compare the terminal velocities of the two spheres.

The lead sphere has a much larger mass, and therefore a larger weight. It must fall at a significantly faster speed to generate enough drag to equal its weight.

A larger downward force (weight) requires a proportionally larger upward force (drag) to achieve equilibrium.

Key Concept

Terminal Velocity and Atmospheric Drag

Estimated Time:1m 30s

Question 25Question →

A construction worker on a tall scaffolding accidentally drops a heavy steel wrench. Assuming there is no air resistance and the acceleration due to gravity is $32\text{ ft/s}^2$ , the wrench hits the ground after exactly $4$ seconds. What is the height from which the wrench was dropped, in feet?

Show answer & explanation

Answer: 256

Answer

The wrench was dropped from a height of 256 feet.

The height of the drop is found by calculating the total distance the wrench falls in 4 seconds under the continuous influence of gravity. Using the standard kinematic equation for an object dropped from rest,

d = \frac{1}{2}gt^2

, and the given gravitational acceleration of

32\text{ ft/s}^2

, the calculation is

0.5 \times 32 \times (4^2) = 16 \times 16 = 256

feet.

Step-by-Step Solution

1

Identify the known variables from the problem statement.

Initial velocity

v_i = 0

, time

t = 4\text{ s}

, gravity

g = 32\text{ ft/s}^2

.

Establishing the knowns is required to select the appropriate kinematic formula for free fall.

2

Select the correct formula for calculating the distance fallen.

d = \frac{1}{2}gt^2

This formula calculates distance when an object accelerates uniformly from rest.

3

Substitute the known values into the equation and perform the calculation.

d = 0.5 \times 32 \times 16 = 256

To compute the final numerical distance.

Key Concept

Calculating distance in free fall using gravitational acceleration

Question 26Question →

An engineer is testing the impact resistance of a $3\text{ kg}$ structural component by releasing it from rest in a drop tower. Ignoring air resistance and assuming the acceleration due to gravity is $9.8\text{ m/s}^2$ , what is the downward velocity of the component exactly $2\text{ seconds}$ after it is released?

9.8\text{ m/s}

19.6\text{ m/s}

29.4\text{ m/s}

58.8\text{ m/s}

Show answer & explanation

Answer: $19.6\text{ m/s}$

Answer

The downward velocity of the component is

19.6\text{ m/s}

.

The velocity of an object in free fall starting from rest is determined by multiplying the acceleration due to gravity by the time elapsed (

v = gt

). Because air resistance is ignored, the

3\text{ kg}

mass is completely irrelevant to the fall speed. Multiplying

9.8\text{ m/s}^2

by

2\text{ seconds}

yields exactly

19.6\text{ m/s}

.

Step-by-Step Solution

1

Identify the given variables needed for the kinematic equation.

Initial velocity (

v_0

) =

0\text{ m/s}

, Acceleration (

g

) =

9.8\text{ m/s}^2

, Time (

t

) =

2\text{ s}

, Mass (

m

) =

3\text{ kg}

.

Extracting the knowns determines which formula is needed to find the final velocity.

2

Select the correct formula for velocity in free fall and eliminate irrelevant information.

v = v_0 + gt

. Mass (

3\text{ kg}

) is not needed.

All objects in free fall accelerate at the same rate regardless of their mass when air resistance is ignored.

3

Substitute the values and calculate the final velocity.

v = 0 + (9.8\text{ m/s}^2 \times 2\text{ s}) = 19.6\text{ m/s}

.

Multiplying the constant acceleration by the time elapsed gives the instantaneous velocity.

Key Concept

In the absence of air resistance, all objects fall with the same constant acceleration regardless of mass, and velocity is the product of this acceleration and time (

v = gt

).

Question 27Question →

A standard shipping crate is measured on Earth (where the acceleration due to gravity is $10 \text{ m/s}^2$ ) using a mechanical balance scale. The balance scale determines that the crate has a mass of $400 \text{ kg}$ . The crate is then transported to a planetary outpost where the local gravitational acceleration is $4 \text{ m/s}^2$ . At the outpost, the crate is hoisted using a heavy-duty spring scale. This spring scale was manufactured and calibrated on Earth to display mass in kilograms based on Earth's gravity. What numerical value will this spring scale display when weighing the crate at the outpost?

Show answer & explanation

Answer: 160

Answer

160

The true mass of the crate is 400 kg, which does not change based on location. On the outpost, the crate's weight (force) is 400 kg × 4 m/s² = 1600 N. The spring scale stretches based entirely on this 1600 N force. Because it was calibrated on Earth (where the acceleration due to gravity is 10 m/s²), its internal mechanism is designed to display '1 kg' for every 10 N of force it supports. Therefore, an applied force of 1600 N causes the scale to point to 1600 / 10 = 160.

Step-by-Step Solution

1

Determine the true mass of the crate.

400 kg

Mass is the amount of matter and is invariant. A balance scale compares masses and measures true mass accurately regardless of the local gravitational field.

2

Calculate the downward force (weight) exerted by the crate on the spring scale at the outpost.

1600 N

Weight depends on local gravity. It is calculated using the formula Weight = mass × local gravity, which is 400 kg × 4 m/s².

3

Determine the calibration factor of the Earth-calibrated spring scale.

10 N per 1 kg of display

A spring scale physically stretches in response to force. Because it was calibrated to display mass on Earth, it assumes that every 10 N of applied force corresponds to 1 kg of mass.

4

Calculate the value displayed by the spring scale.

160

The scale converts the 1600 N force it physically experiences into a kilogram reading using its Earth calibration: 1600 N / 10 m/s² = 160.

Key Concept

Mass is an invariant property of matter, while weight is the localized gravitational force acting on that mass. A spring scale measures force (weight) rather than mass, meaning a spring scale calibrated in kilograms on Earth will display incorrect mass readings in different gravitational fields.

Question 28Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are determined by dividing the net force by the mass for each scenario: the aerospace capsule accelerates at

3.0 \text{ m/s}^2

, the forklift accelerates at

1.5 \text{ m/s}^2

, the bridge section accelerates at

0.5 \text{ m/s}^2

, and the submarine ballast accelerates at

4.0 \text{ m/s}^2

.

Newton's Second Law dictates that acceleration is directly proportional to net force and inversely proportional to mass (

a = F/m

). By dividing the net force by the mass in each of the four scenarios, the correct acceleration values are directly calculated and paired.

Step-by-Step Solution

1

Identify the known variables provided in each scenario.

Each scenario provides the net force (

F

) in Newtons and the object's mass (

m

) 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 = ma

) to solve for the missing variable.

The formula becomes

a = F/m

.

Because the right items represent acceleration, the equation must be isolated for acceleration (

a

) before computing.

3

Calculate the specific acceleration for each mass and force pairing.

Capsule:

3,600 / 1,200 = 3.0 \text{ m/s}^2

. Forklift:

2,250 / 1,500 = 1.5 \text{ m/s}^2

. Bridge section:

1,250 / 2,500 = 0.5 \text{ m/s}^2

. Ballast:

16,000 / 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.

Key Concept

Solving for acceleration using Newton's Second Law (

a = F/m

)

Question 29Question →

A $500\text{ kg}$ steel block is dropped from rest into a deep vertical mineshaft. Ignoring air resistance, arrange the following one-second time intervals in order according to the distance the block travels during each specific interval, from the shortest distance to the longest distance.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence is chronological: the 1st second, followed by the 2nd second, 3rd second, and 4th second.

In free fall, an object undergoes constant acceleration due to gravity. Because its downward speed is continuously increasing, the distance it travels during each successive equal time interval also increases. The block's heavy mass is a distracter; it does not change the fact that the shortest distance is covered in the first second (when it is slowest), and the longest distance is covered in the final second.

Step-by-Step Solution

1

Analyze the relationship between gravity, acceleration, and the speed of a falling object.

Gravity causes a constant downward acceleration on the block, meaning its speed increases steadily as time passes.

Understanding constant acceleration is required to determine how distance changes over equal time increments.

2

Determine the distance traveled during each successive one-second interval.

Since the block's average speed is higher during each subsequent second, the distance it travels in each one-second interval must also be strictly greater than the previous interval.

Distance equals average speed multiplied by time. A higher average speed over the same one-second duration yields a longer distance.

3

Order the time intervals from the shortest distance to the longest distance.

The 1st second has the lowest average speed (shortest distance), followed in order by the 2nd, 3rd, and 4th seconds (longest distance).

This sequences the intervals properly according to the prompt's instructions.

Key Concept

Acceleration due to gravity causes a freely falling object's speed to continuously increase, resulting in strictly greater distances covered during each successive time interval, regardless of its mass.

Question 30Question →

An aerospace technician is testing an experimental probe with a weight of $50\text{ N}$ on Earth (where $g = 10\text{ m/s}^2$ ). The probe is released from rest inside an evacuated drop tube, ensuring there is no air resistance. It falls freely for $200\text{ milliseconds}$ . What is the probe's velocity right before impact, and how would this impact velocity change if a $100\text{ N}$ probe were tested under the exact same conditions?

2000\text{ m/s}

; the impact velocity would remain exactly the same because free-fall acceleration is independent of an object's weight.

1.0\text{ m/s}

; the impact velocity would increase to

2.0\text{ m/s}

because the object's mass, rather than a constant gravitational rate, determines its falling speed.

2.0\text{ m/s}

; the impact velocity would remain exactly the same because free-fall acceleration is independent of an object's weight.

2.0\text{ m/s}

; the impact velocity would increase to

4.0\text{ m/s}

because a greater mass results in a proportionally greater rate of acceleration.

Show answer & explanation

Answer: $2.0\text{ m/s}$ ; the impact velocity would remain exactly the same because free-fall acceleration is independent of an object's weight.

Answer

The probe's impact velocity is

2.0\text{ m/s}

, and the velocity would remain exactly the same for the heavier probe.

First, the fall time must be converted to standard units (

200\text{ ms} = 0.2\text{ s}

). Using the velocity formula

v = g \times t

, the speed is calculated as

10 \times 0.2 = 2.0\text{ m/s}

. Second, in the complete absence of air resistance, the acceleration due to gravity is a constant. Therefore, a heavier

100\text{ N}

probe would fall at the exact same rate and strike the ground with the exact same velocity.

Step-by-Step Solution

1

Convert the free-fall time from milliseconds to seconds.

200\text{ ms} = 0.2\text{ seconds}

Standard kinematic equations require time to be in standard SI units (seconds) to match the acceleration rate (

10\text{ m/s}^2

).

2

Calculate the impact velocity using the formula for an object falling from rest (

v = g \times t

).

v = 10\text{ m/s}^2 \times 0.2\text{ s} = 2.0\text{ m/s}

This determines the exact speed of the

50\text{ N}

probe right before it strikes the bottom of the drop tube.

3

Evaluate how doubling the probe's weight to

100\text{ N}

affects the velocity.

The impact velocity remains

2.0\text{ m/s}

.

In an evacuated space (a vacuum), there is no air resistance. All objects accelerate due to gravity at the same constant rate, regardless of their individual masses or weights.

Key Concept

Free Fall Velocity and Mass Independence

Estimated Time:1m 0s

Question 31Question →

An engineer on an unknown planet evaluates a solid metallic crate using two different instruments. First, the crate perfectly balances on a double-pan beam balance against standard reference masses totaling $150\text{ kg}$ . Second, the crate is suspended from a calibrated spring scale, which gives a reading of $600\text{ N}$ . If the crate is transported back to Earth (where gravitational acceleration is $10\text{ m/s}^2$ ), what will the spring scale and the beam balance read, respectively?

Spring scale:

1500\text{ N}

; Beam balance:

150\text{ kg}

Spring scale:

600\text{ N}

; Beam balance:

150\text{ kg}

Spring scale:

1500\text{ N}

; Beam balance:

375\text{ kg}

Spring scale:

150\text{ N}

; Beam balance:

1500\text{ kg}

Show answer & explanation

Answer: Spring scale: $1500\text{ N}$ ; Beam balance: $150\text{ kg}$

Answer

The spring scale will read

1500\text{ N}

and the beam balance will read

150\text{ kg}

.

The correct answer accurately distinguishes between mass and weight. Mass is the amount of matter and does not change regardless of location, meaning the beam balance will still read

150\text{ kg}

on Earth. Weight is the force of gravity acting on that mass. A spring scale measures this downward force. On Earth, the force is

W = mg = 150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the crate using the beam balance reading.

Mass (

m

) =

150\text{ kg}

.

A double-pan beam balance compares unknown mass against known reference masses. Because local gravity acts equally on both pans, gravity cancels out, meaning the balance directly measures true mass, which is invariant.

2

Calculate the weight of the crate on Earth.

Weight on Earth (

W

) =

1500\text{ N}

.

Weight is the gravitational force exerted on an object, calculated using

W = mg

. Using Earth's gravity (

g = 10\text{ m/s}^2

), the weight is

150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}

.

3

Determine what each instrument will read on Earth.

Spring scale reads

1500\text{ N}

; Beam balance reads

150\text{ kg}

.

The spring scale measures the downward force of weight (

1500\text{ N}

), while the beam balance continues to measure the invariant mass (

150\text{ kg}

).

Key Concept

Mass is an invariant property of an object measured by a balance, while weight is a force dependent on local gravity measured by a spring scale.

Alternative Method

You can verify the unknown planet's gravity (

g = W/m = 600\text{ N} / 150\text{ kg} = 4\text{ m/s}^2

) to confirm the initial readings are consistent, then reapply the logic to Earth (

g = 10\text{ m/s}^2

) to find the new weight.

Estimated Time:1m 30s

Question 32Question →

Two astronauts are conducting a repair mission in a zero-gravity vacuum environment. A massive 800 kg communication satellite is drifting in a straight line at a constant velocity of 1 m/s. An astronaut maneuvers in front of the satellite and attempts to stop it by pushing against it with her hands. She finds it extremely difficult to bring the satellite to a halt.

Based on the principles of mechanics, which of the following best explains why the weightless satellite is so difficult to stop?

The satellite's significant mass provides a high degree of inertia, which strongly resists any changes to its constant velocity.

The satellite's constant motion generates an active inertial force that continuously pushes against the astronaut to maintain its speed.

The satellite is in a zero-gravity environment but retains its heavy weight, which creates the physical resistance the astronaut feels when pushing.

The astronaut's push creates an equal and opposite reaction force from the satellite that cancels out her attempt to decelerate it.

Show answer & explanation

Answer: The satellite's significant mass provides a high degree of inertia, which strongly resists any changes to its constant velocity.

Answer

The satellite's significant mass provides a high degree of inertia, which strongly resists any changes to its constant velocity.

According to Newton's First Law, inertia is the property of an object that resists any change to its velocity. The amount of inertia an object possesses is directly proportional to its mass, regardless of gravitational pull. Therefore, an 800 kg satellite requires a significant force (or a prolonged period of time) to decelerate, purely because of its high mass and the resulting inertia.

Step-by-Step Solution

1

Analyze the environmental conditions described in the scenario.

The scenario takes place in a zero-gravity vacuum, meaning there is no weight and no air resistance acting on the satellite.

Understanding the environment helps eliminate forces like gravity and friction as contributing factors to the resistance felt by the astronaut.

2

Identify the relevant mechanical property of the satellite.

Although the satellite is weightless, it has a mass of 800 kg.

Mass is an intrinsic property of matter that remains constant regardless of gravitational pull.

3

Apply Newton's First Law of Motion to the situation.

The satellite's large mass means it has high inertia, which naturally resists any acceleration or deceleration.

Inertia is the tendency of an object to resist changes in its state of motion, and it is directly proportional to an object's mass.

Key Concept

Newton's First Law (Inertia)

Question 33Question →

A driver must execute an emergency stop on a dry asphalt road. The vehicle is equipped with an anti-lock braking system (ABS) that uses threshold braking, allowing the tires to continue rotating just on the verge of locking up. If the ABS were disabled and the driver fully locked the brakes, causing the tires to skid continuously across the surface, how would the frictional resistance between the tires and the road change?

The resistance would decrease, because the skidding tires rely on kinetic friction, which is generally weaker than the maximum static friction maintained during threshold braking.

The resistance would increase, because the skidding tires deform and flatten, creating a larger contact surface area with the road.

The resistance would remain unchanged, because frictional force depends on the vehicle's mass and is completely unaffected by its gravitational weight or rotation.

The resistance would increase, because the forward inertia of the skidding vehicle creates a larger opposing reaction force from the asphalt.

Show answer & explanation

Answer: The resistance would decrease, because the skidding tires rely on kinetic friction, which is generally weaker than the maximum static friction maintained during threshold braking.

Answer

The resistance would decrease, because the skidding tires rely on kinetic friction, which is generally weaker than the maximum static friction maintained during threshold braking.

As long as a tire is rolling without slipping (as is maintained during threshold braking), the exact point of the tire touching the road is momentarily stationary relative to the ground. This means it utilizes static friction. Once the brakes lock and the tire skids, the rubber slides across the surface, switching the force to kinetic friction. Because the coefficient of static friction is greater than the coefficient of kinetic friction, a sliding tire actually provides less overall stopping resistance than a tire rolling on the verge of locking.

Step-by-Step Solution

1

Identify the type of friction present during threshold braking when the tires are still rotating.

The bottom of a rolling tire does not slide against the road, meaning the system utilizes static friction at the contact patch.

Understanding the physical state of rolling without slipping is necessary to determine the active frictional forces.

2

Identify the type of friction present during a full brake lockup (skid).

When the tires lock, they slide across the asphalt, meaning the system transitions to utilizing kinetic (sliding) friction.

Skidding fundamentally changes the interaction between the surfaces from gripping to sliding.

3

Compare the magnitudes of the relevant friction coefficients.

For standard materials like rubber on asphalt, the coefficient of static friction is greater than the coefficient of kinetic friction.

This physical rule dictates which braking method provides the most stopping force.

4

Determine the net effect on frictional resistance.

Because kinetic friction is weaker than maximum static friction, locking the brakes and skidding results in decreased frictional resistance.

Combining the physical state with the friction coefficients provides the final answer.

Key Concept

Static versus Kinetic Friction in Rolling Mechanisms

Question 34Question →

A steel construction beam has a mass of $120 \text{ kg}$ on Earth, where the gravitational acceleration is approximately $10 \text{ m/s}^2$ . The beam is subsequently transported to a lunar outpost, where the local gravitational acceleration is $1.6 \text{ m/s}^2$ . What is the mass of the beam on the lunar outpost?

19.2 \text{ kg}

192 \text{ kg}

120 \text{ kg}

192 \text{ N}

Show answer & explanation

Answer: $120 \text{ kg}$

Answer

120 \text{ kg}

The correct answer is

120 \text{ kg}

because mass represents the physical amount of matter in the steel beam. This property is invariant and does not change when moved to a different gravitational environment, unlike weight.

Step-by-Step Solution

1

Identify the quantity requested in the problem.

The question specifically asks for the mass of the beam on the Moon.

Carefully distinguishing between requested mass and weight prevents calculation errors based on gravity.

2

Recall the definition and properties of mass.

Mass is the amount of matter in an object and is an invariant property, meaning it does not change based on location.

Unlike weight, which is a force dependent on the local gravitational acceleration (

W = mg

), mass remains constant universally.

3

Determine the mass on the Moon based on its initial mass.

Since the mass on Earth is

120 \text{ kg}

, the mass on the Moon remains exactly

120 \text{ kg}

.

The change in gravitational acceleration from

10 \text{ m/s}^2

to

1.6 \text{ m/s}^2

affects only the weight of the object, not its mass.

Key Concept

Mass is an intrinsic, invariant property of an object (measured in kilograms), whereas weight is the location-dependent gravitational force exerted on that mass (measured in Newtons).

Question 35Question →

A special effects crew is dropping various props into a deep gorge for a movie scene. Assuming air resistance is negligible, arrange the following drops in order from the shortest time in the air to the longest time in the air before hitting the bottom.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from shortest to longest fall time is: the 100 kg safe (5 m), the 2 kg camera (20 m), the 50 kg sandbag (45 m), and finally the 10 kg stunt dummy (80 m).

The time an object spends in free fall from rest is determined entirely by the height from which it is dropped, according to the kinematic equation

t = \sqrt{\frac{2d}{g}}

. Because air resistance is negligible in this scenario, the mass of the objects is completely irrelevant. Therefore, arranging the objects from shortest to longest fall time is identical to arranging them from lowest to highest drop height:

5\text{ m}

(safe),

20\text{ m}

(camera),

45\text{ m}

(sandbag), and

80\text{ m}

(dummy).

Step-by-Step Solution

1

Identify the relevant formula for the time of free fall from rest.

d = \frac{1}{2}gt^2

, which rearranges to

t = \sqrt{\frac{2d}{g}}

.

To determine which variables affect the time an object spends in the air.

2

Analyze the role of mass in the free fall equation.

Mass (

m

) does not appear in the equation.

In the absence of air resistance, all objects accelerate toward Earth at the exact same rate (

g

), regardless of how heavy they are.

3

Compare the heights from which the objects are dropped.

The heights are

5\text{ m}

,

20\text{ m}

,

45\text{ m}

, and

80\text{ m}

.

Since fall time is entirely dependent on the initial height (

d

), the object dropped from the lowest height hits the ground first.

4

Order the objects based on their drop heights from shortest to highest.

Safe (

5\text{ m}

) < Camera (

20\text{ m}

) < Sandbag (

45\text{ m}

) < Dummy (

80\text{ m}

).

A shorter fall distance results in a shorter amount of time spent in the air.

Key Concept

In a vacuum or when air resistance is negligible, the acceleration due to gravity is constant for all objects. Therefore, the time it takes to fall depends only on the initial height, not the mass.

Question 36Question →

A newly designed lunar rover is being tested in a facility on Earth. The rover has a measured downward weight of $4500 \text{ N}$ . Assuming the acceleration due to gravity on Earth is $10 \text{ m/s}^2$ , what is the mass of the rover in kilograms?

Show answer & explanation

Answer: 450

Answer

The mass of the rover is 450 kg.

Mass represents the amount of matter in an object and is found by dividing its weight (the force of gravity acting upon it) by the local gravitational acceleration. Using the formula

m = \frac{W}{g}

, substituting

4500 \text{ N}

for

W

and

10 \text{ m/s}^2

for

g

yields exactly

450 \text{ kg}

.

Step-by-Step Solution

1

Identify the given variables for weight and gravity.

W = 4500 \text{ N}

,

g = 10 \text{ m/s}^2

These are the necessary components to determine the object's mass.

2

State the relationship between weight and mass.

W = m \times g

This is Newton's Second Law applied to the force of gravity.

3

Rearrange the formula to isolate mass (

m

).

m = \frac{W}{g}

We need to solve for the unknown variable, mass.

4

Substitute the values and divide.

m = \frac{4500}{10} = 450

Performing the arithmetic gives the amount of matter inside the rover.

Key Concept

Weight is the gravitational force exerted on an object, calculated as the product of its mass and local gravitational acceleration (

W = mg

). Mass is an invariant property that can be found by dividing weight by gravity.

Estimated Time:45s

Question 37Question →

Engineers are conducting a drop test in a massive vertical vacuum chamber, entirely eliminating air drag. They simultaneously release a robust data-recording payload with a mass of $250\text{ kg}$ and a small calibration weight with a mass of $10\text{ kg}$ from the exact same altitude. Which of the following statements correctly describes the forces and motion involved in this scenario?

The data-recording payload experiences a greater gravitational force than the calibration weight, but both objects accelerate downward at the same rate.

The data-recording payload experiences a greater gravitational force, causing it to accelerate downward faster than the calibration weight.

Both objects experience the exact same gravitational force because they are in a vacuum, resulting in equal downward acceleration.

The calibration weight accelerates downward faster because its lower mass provides less inertial resistance to the vacuum environment.

Show answer & explanation

Answer: The data-recording payload experiences a greater gravitational force than the calibration weight, but both objects accelerate downward at the same rate.

Answer

The correct answer states that the heavier payload experiences greater gravitational force, but both objects share the same downward acceleration.

The gravitational force (often called weight) acting on an object is calculated by multiplying its mass by the acceleration due to gravity (

F = mg

). Because the payload has a larger mass, it experiences a significantly larger gravitational force. However, an object's resistance to changes in motion (inertia) is also determined by its mass. When we calculate acceleration using Newton's Second Law (

a = F/m

), we substitute

mg

for

F

, resulting in

a = mg/m

. The mass cancels out, proving that both objects, regardless of how heavy they are, accelerate downward at the exact same rate (approximately

9.8\text{ m/s}^2

) when air resistance is absent.

Step-by-Step Solution

1

Identify the primary forces acting on the objects during the drop.

The only force acting on both objects is the gravitational force (weight).

The scenario specifies a vacuum chamber, which means air resistance (drag) is zero.

2

Compare the magnitude of the gravitational force on each object.

The

250\text{ kg}

payload experiences 25 times more gravitational force than the

10\text{ kg}

weight.

Gravitational force is calculated using

F = mg

. Since gravitational acceleration (

g

) is constant, the force is directly proportional to the object's mass.

3

Determine the acceleration of each object using Newton's Second Law.

Both objects accelerate downward at the exact same rate (

g

).

Acceleration is force divided by mass (

a = F/m

). Because the gravitational force

F

increases proportionally with mass

m

, the ratio remains perfectly constant for all objects.

Key Concept

Gravitational Force vs. Acceleration in Free Fall

Question 38Question →

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

4.0 \text{ m/s}^2

5.0 \text{ m/s}^2

6.0 \text{ m/s}^2

0.5 \text{ m/s}^2

Show answer & explanation

Answer: $5.0 \text{ m/s}^2$

Answer

The new acceleration is

5.0 \text{ m/s}^2

.

According to Newton's Second Law, acceleration is determined by the net force acting on an object (

a = \frac{F_{net}}{m}

). The net force is the forward thrust minus the friction. With a new thrust of

12,000 \text{ N}

and a constant friction of

2,000 \text{ N}

, the net force is

10,000 \text{ N}

. Dividing this by the mass of

2,000 \text{ kg}

yields an acceleration of

5.0 \text{ m/s}^2

.

Step-by-Step Solution

1

Identify the forces acting on the vehicle in the new scenario.

Forward thrust (

F_{thrust}

) =

12,000 \text{ N}

. Frictional resistance (

F_{friction}

) =

2,000 \text{ N}

.

Newton's Second Law requires the net force, which is the sum of all opposing forces.

2

Calculate the new net force.

F_{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 = ma

) to find the new acceleration.

a = \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.

Key Concept

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.

Question 39Question →

An unsecured heavy toolbox is resting in the bed of a parked pickup truck. If the truck suddenly accelerates forward in a straight line, what will initially happen to the toolbox and why?

It will slide toward the rear of the truck because its inertia resists the change in motion, keeping it initially at rest.

It will slide toward the front of the truck because a reaction force pulls it in the direction of the truck's acceleration.

It will slide toward the rear of the truck because an active backward force is created to balance the truck's forward motion.

It will slide toward the front of the truck because the toolbox's weight is transferred horizontally when the vehicle moves.

Show answer & explanation

Answer: It will slide toward the rear of the truck because its inertia resists the change in motion, keeping it initially at rest.

Answer

The toolbox will slide toward the rear of the truck because its inertia resists the change in motion, keeping it initially at rest.

According to Newton's First Law (inertia), an object at rest will stay at rest unless acted upon by an outside force. Because the toolbox is unsecured, the truck's forward force is not directly transferred to it. The toolbox tends to stay in its original position relative to the ground while the truck bed moves forward, resulting in the toolbox sliding toward the rear of the truck.

Step-by-Step Solution

1

Analyze the initial state of the toolbox.

The toolbox is at rest relative to the ground.

Newton's First Law applies to objects at rest, stating they will remain at rest unless an unbalanced force is applied.

2

Determine the effect of the truck's sudden forward acceleration.

The truck bed moves forward, but because the toolbox is unsecured, the forward force is not fully transferred to the toolbox.

Objects require a direct applied force to change their state of motion.

3

Conclude the toolbox's motion relative to the truck.

The toolbox tends to remain in its original position relative to the ground while the truck bed moves forward underneath it, making it appear to slide toward the rear tailgate.

This is a direct demonstration of inertia—the tendency of an object to resist changes in its motion.

Key Concept

Newton's First Law (Inertia)

Estimated Time:45s

Question 40Question →

An automotive engineer is designing a mechanical disc brake system to stop a heavy rotating axle. The engineer is evaluating two brake pad designs made of the exact same friction material: Design X utilizes a single large brake pad, while Design Y utilizes a smaller brake pad with exactly half the surface area of Design X. During a braking event, the hydraulic system applies an identical clamping force (normal force) to the pad in both designs. Assuming the coefficient of kinetic friction remains constant, which of the following best describes the resulting kinetic frictional resistance and the reasoning behind it?

Design Y produces significantly less kinetic frictional resistance because its smaller contact area interacts with fewer surface imperfections on the brake disc.

Both designs produce the same kinetic frictional resistance because the total frictional force depends on the applied clamping force and the coefficient of friction, regardless of the contact area.

Design X produces greater kinetic frictional resistance because the larger pad contains more mass, which converts the identical hydraulic pressure into a heavier total braking force.

Design Y produces more kinetic frictional resistance because concentrating the hydraulic force into a smaller area increases the pad's inertia, multiplying the reaction force.

Show answer & explanation

Answer: Both designs produce the same kinetic frictional resistance because the total frictional force depends on the applied clamping force and the coefficient of friction, regardless of the contact area.

Answer

Both brake pad designs produce the identical amount of kinetic frictional resistance because macroscopic friction is independent of the contact surface area.

The correct answer accurately applies the standard physical model of kinetic friction, where the total frictional force (

F_k

) is solely the product of the coefficient of kinetic friction (

\mu_k

) and the normal force (

F_n

). Because both brake pad designs are made of identical material and are subjected to the exact same hydraulic clamping force, the resulting kinetic friction is identical. In solid mechanics, sliding friction does not depend on the macroscopic surface area in contact.

Step-by-Step Solution

1

Identify the primary equation used to calculate kinetic friction.

F_k = \mu_k \times F_n

This formula establishes that kinetic frictional force (

F_k

) depends strictly on the coefficient of kinetic friction (

\mu_k

) and the normal force (

F_n

).

2

Evaluate the variables given in the scenario for both designs.

Both designs possess the same normal force (identical clamping force) and the same coefficient of friction (exact same friction material).

The prompt explicitly states the hydraulic system applies an identical clamping force and the materials are the same.

3

Determine the impact of the differing surface areas on the total frictional force.

The surface area does not affect the calculation. Both designs yield the exact same frictional resistance.

Surface area is not a variable in the standard kinetic friction formula; increased area distributes the force resulting in lower pressure, while decreased area concentrates the force resulting in higher pressure, keeping total friction constant.

Key Concept

Independence of kinetic friction from surface area

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