Mass versus Weight

9 questions

Question 1Question →

A specialized rover is designed to operate on both Earth (where gravitational acceleration is 10 m/s210\text{ m/s}^2) and an exoplanet (where gravitational acceleration is 15 m/s215\text{ m/s}^2). The rover carries a scientific payload that is measured to have a weight of 600 N600\text{ N} on Earth. When the rover arrives on the exoplanet, it suspends the payload from a spring scale to verify its properties, and later pushes the payload horizontally across a frictionless ice field to accelerate it at 4 m/s24\text{ m/s}^2. What will the spring scale read, and what horizontal force must the rover apply?

Show answer & explanation

Answer: Spring scale: 900 N900\text{ N}; Horizontal force: 240 N240\text{ N}

Answer

Spring scale: 900 N900\text{ N}; Horizontal force: 240 N240\text{ N}
The payload's mass is an invariant property, calculated from its Earth weight (m=Wg=60010=60 kgm = \frac{W}{g} = \frac{600}{10} = 60\text{ kg}). On the exoplanet, the spring scale measures the local gravitational force (weight), which is mass times local gravity (60×15=900 N60 \times 15 = 900\text{ N}). The horizontal force depends only on the invariant mass and the desired acceleration (F=ma=60×4=240 NF = ma = 60 \times 4 = 240\text{ N}).

Step-by-Step Solution

1
Determine the invariant mass of the payload using its Earth weight.
m=WEarthgEarth=600 N10 m/s2=60 kgm = \frac{W_{\text{Earth}}}{g_{\text{Earth}}} = \frac{600\text{ N}}{10\text{ m/s}^2} = 60\text{ kg}
Mass is the amount of matter and remains constant regardless of location. It is required to determine both the new weight and the horizontal force.
2
Calculate the new weight on the exoplanet to find the spring scale reading.
Wexo=m×gexo=60 kg×15 m/s2=900 NW_{\text{exo}} = m \times g_{\text{exo}} = 60\text{ kg} \times 15\text{ m/s}^2 = 900\text{ N}
A spring scale measures the local force of gravity (weight), which depends on the local gravitational acceleration.
3
Calculate the required horizontal force using Newton's Second Law.
F=m×a=60 kg×4 m/s2=240 NF = m \times a = 60\text{ kg} \times 4\text{ m/s}^2 = 240\text{ N}
Horizontal acceleration on a frictionless surface depends only on the object's mass (inertia) and the applied force, not on its weight.

Key Concept

Mass is an invariant property that determines inertia and relates to force (F=maF=ma), while weight is a location-dependent gravitational force (W=mgW=mg) measured by spring scales.
Question 2Question →

A standard shipping crate is measured on Earth (where the acceleration due to gravity is 10 m/s210 \text{ m/s}^2) using a mechanical balance scale. The balance scale determines that the crate has a mass of 400 kg400 \text{ kg}. The crate is then transported to a planetary outpost where the local gravitational acceleration is 4 m/s24 \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?

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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 3Question →

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 kg150\text{ kg}. Second, the crate is suspended from a calibrated spring scale, which gives a reading of 600 N600\text{ N}. If the crate is transported back to Earth (where gravitational acceleration is 10 m/s210\text{ m/s}^2), what will the spring scale and the beam balance read, respectively?

Show answer & explanation

Answer: Spring scale: 1500 N1500\text{ N}; Beam balance: 150 kg150\text{ kg}

Answer

The spring scale will read 1500 N1500\text{ N} and the beam balance will read 150 kg150\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 kg150\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 kg×10 m/s2=1500 NW = 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 (mm) = 150 kg150\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 (WW) = 1500 N1500\text{ N}.
Weight is the gravitational force exerted on an object, calculated using W=mgW = mg. Using Earth's gravity (g=10 m/s2g = 10\text{ m/s}^2), the weight is 150 kg×10 m/s2=1500 N150\text{ kg} \times 10\text{ m/s}^2 = 1500\text{ N}.
3
Determine what each instrument will read on Earth.
Spring scale reads 1500 N1500\text{ N}; Beam balance reads 150 kg150\text{ kg}.
The spring scale measures the downward force of weight (1500 N1500\text{ N}), while the beam balance continues to measure the invariant mass (150 kg150\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 N/150 kg=4 m/s2g = 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 m/s2g = 10\text{ m/s}^2) to find the new weight.
Estimated Time:1m 30s
Question 4Question →

A steel construction beam has a mass of 120 kg120 \text{ kg} on Earth, where the gravitational acceleration is approximately 10 m/s210 \text{ m/s}^2. The beam is subsequently transported to a lunar outpost, where the local gravitational acceleration is 1.6 m/s21.6 \text{ m/s}^2. What is the mass of the beam on the lunar outpost?

Show answer & explanation

Answer: 120 kg120 \text{ kg}

Answer

120 kg120 \text{ kg}
The correct answer is 120 kg120 \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=mgW = mg), mass remains constant universally.
3
Determine the mass on the Moon based on its initial mass.
Since the mass on Earth is 120 kg120 \text{ kg}, the mass on the Moon remains exactly 120 kg120 \text{ kg}.
The change in gravitational acceleration from 10 m/s210 \text{ m/s}^2 to 1.6 m/s21.6 \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 5Question →

A newly designed lunar rover is being tested in a facility on Earth. The rover has a measured downward weight of 4500 N4500 \text{ N}. Assuming the acceleration due to gravity on Earth is 10 m/s210 \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=Wgm = \frac{W}{g}, substituting 4500 N4500 \text{ N} for WW and 10 m/s210 \text{ m/s}^2 for gg yields exactly 450 kg450 \text{ kg}.

Step-by-Step Solution

1
Identify the given variables for weight and gravity.
W=4500 NW = 4500 \text{ N}, g=10 m/s2g = 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×gW = m \times g
This is Newton's Second Law applied to the force of gravity.
3
Rearrange the formula to isolate mass (mm).
m=Wgm = \frac{W}{g}
We need to solve for the unknown variable, mass.
4
Substitute the values and divide.
m=450010=450m = \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=mgW = mg). Mass is an invariant property that can be found by dividing weight by gravity.
Estimated Time:45s
Question 6Question →

Match each physical concept or measuring instrument to its correct characteristic or function regarding mass and weight.

Click a left item, then click its matching right item

Items

Mass
Weight
Double-pan balance
Spring scale

Matches

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Answer

Mass pairs with the constant fundamental property; Weight pairs with the gravity-dependent force; Double-pan balance pairs with the consistent standard-comparison device; Spring scale pairs with the coil-based device.
The correct pairings accurately reflect that mass is an invariant scalar quantity measured consistently by a balance, while weight is a vector force that depends on local gravity and is measured by a spring scale.

Step-by-Step Solution

1
Distinguish between the conceptual definitions of mass and weight.
Mass is identified as a constant amount of matter, whereas weight is the variable force of gravity acting on that matter.
To correctly match the physical properties with their fundamental definitions.
2
Determine the operational principles of the two measuring instruments.
A double-pan balance compares masses directly and works consistently regardless of local gravity, whereas a spring scale measures downward force (weight) via coil stretch, which varies with gravity.
To correctly pair the measuring devices with their functional characteristics across different gravitational fields.

Key Concept

Mass is an invariant property measured by comparing it to known standards (using a balance), while weight is a location-dependent force measured by its effect on a spring (using a spring scale).
Question 7Question →

A mining crew extracts a dense mineral boulder. In their vessel's main processing bay, where artificial gravity is maintained at 10 m/s210 \text{ m/s}^2, a spring scale indicates the boulder has a weight of 1500 N1500 \text{ N}. The boulder is then transported to a storage facility on a small moon where the local gravitational acceleration is 2 m/s22 \text{ m/s}^2. What are the mass and weight of the boulder at the lunar storage facility?

Show answer & explanation

Answer: Mass: 150 kg150 \text{ kg}, Weight: 300 N300 \text{ N}

Answer

The mass is 150 kg150 \text{ kg} and the weight is 300 N300 \text{ N}.
The correct answer accurately calculates the initial mass as 150 kg150 \text{ kg} using the weight on the vessel (1500 N/10 m/s21500 \text{ N} / 10 \text{ m/s}^2). Because mass is an invariant property, it remains 150 kg150 \text{ kg} on the moon. The new weight is then found by multiplying this mass by the moon's local gravity (150 kg×2 m/s2=300 N150 \text{ kg} \times 2 \text{ m/s}^2 = 300 \text{ N}).

Step-by-Step Solution

1
Calculate the mass of the boulder using its initial weight and the artificial gravity in the processing bay.
m=W1g1=1500 N10 m/s2=150 kgm = \frac{W_1}{g_1} = \frac{1500 \text{ N}}{10 \text{ m/s}^2} = 150 \text{ kg}.
Mass is an intrinsic property that must be determined first to calculate weight in other environments.
2
Determine the mass of the boulder at the lunar storage facility.
The mass remains 150 kg150 \text{ kg}.
Mass is an invariant measure of the amount of matter and does not change with location or local gravity.
3
Calculate the new weight of the boulder using the local gravitational acceleration on the small moon.
W2=m×g2=150 kg×2 m/s2=300 NW_2 = m \times g_2 = 150 \text{ kg} \times 2 \text{ m/s}^2 = 300 \text{ N}.
Weight is the gravitational force acting on an object, which depends on the local acceleration due to gravity.

Key Concept

Mass is an intrinsic, invariant property representing the amount of matter, whereas weight is the location-dependent gravitational force acting upon that mass.
Question 8Question →

An automated logistics drone is transporting a sealed supply crate. On Earth, where the gravitational acceleration is considered to be 10 m/s210 \text{ m/s}^2, the crate has a measured weight of 1400 N1400 \text{ N}. The crate is then delivered to a controlled low-gravity research habitat where the local gravitational acceleration is maintained at 3.5 m/s23.5 \text{ m/s}^2.

What is the weight of the crate inside the low-gravity habitat in Newtons?

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Answer: 490

Answer

490
To find the new weight, first determine the crate's mass, which does not change. Dividing the Earth weight (1400 N1400 \text{ N}) by Earth's gravity (10 m/s210 \text{ m/s}^2) gives a mass of 140 kg140 \text{ kg}. Then, multiply this invariant mass by the habitat's gravity (3.5 m/s23.5 \text{ m/s}^2) to find the new weight: 140 kg×3.5 m/s2=490 N140 \text{ kg} \times 3.5 \text{ m/s}^2 = 490 \text{ N}.

Step-by-Step Solution

1
Calculate the mass of the crate on Earth.
m=1400 N10 m/s2=140 kgm = \frac{1400 \text{ N}}{10 \text{ m/s}^2} = 140 \text{ kg}
Mass is an intrinsic property of the object that remains constant regardless of location.
2
Calculate the weight of the crate in the low-gravity habitat.
W=140 kg×3.5 m/s2=490 NW = 140 \text{ kg} \times 3.5 \text{ m/s}^2 = 490 \text{ N}
Weight is the force exerted by gravity on a mass, which changes depending on the local gravitational acceleration.

Key Concept

Mass is invariant, while weight depends on the local gravitational field.
Question 9Question →

At an Earth-based testing facility (where g=10 m/s2g = 10 \text{ m/s}^2), a suspended spring scale reads 750 N750 \text{ N} when a communications satellite module is attached to it. The module is later transported to a planetary outpost where the local gravitational acceleration is 4 m/s24 \text{ m/s}^2. Which of the following correctly identifies the module's mass and its new weight at the outpost?

Show answer & explanation

Answer: Mass = 75 kg75 \text{ kg}, Weight = 300 N300 \text{ N}

Answer

Mass = 75 kg75 \text{ kg}, Weight = 300 N300 \text{ N}
The correct answer accurately separates the concepts of mass and weight. First, the mass is found by dividing the Earth weight by Earth's gravity (750 N/10 m/s2=75 kg750 \text{ N} / 10 \text{ m/s}^2 = 75 \text{ kg}). Because mass never changes regardless of location, it stays 75 kg75 \text{ kg}. The new weight is then found by multiplying this constant mass by the outpost's gravity (75 kg×4 m/s2=300 N75 \text{ kg} \times 4 \text{ m/s}^2 = 300 \text{ N}).

Step-by-Step Solution

1
Calculate the module's mass using its weight on Earth.
m=Wg=750 N10 m/s2=75 kgm = \frac{W}{g} = \frac{750 \text{ N}}{10 \text{ m/s}^2} = 75 \text{ kg}.
Weight is the product of mass and gravitational acceleration (W=mgW = mg). Rearranging this formula allows us to find the intrinsic mass.
2
Determine the module's mass at the planetary outpost.
The mass remains 75 kg75 \text{ kg}.
Mass is a measure of the amount of matter in an object. It is an invariant property that does not change when moved to a different gravitational field.
3
Calculate the module's new weight at the planetary outpost.
Wnew=m×goutpost=75 kg×4 m/s2=300 NW_{new} = m \times g_{outpost} = 75 \text{ kg} \times 4 \text{ m/s}^2 = 300 \text{ N}.
Weight is a force that depends on the local gravitational pull. The new weight must be calculated using the outpost's specific gravitational acceleration.

Key Concept

Mass is an invariant property representing the amount of matter (measured in kilograms), whereas weight is the gravitational force acting on that mass (measured in Newtons) and varies depending on the local gravitational field.
Mass versus Weight Practice Questions — Armed Services Vocational Aptitude Battery (ASVAB) | Examkin