Measurement of Mass and Weight Practice Questions — JAMB UTME

Question 1Question →

A body of mass $50\text{ kg}$ is suspended from a spring balance attached to the ceiling of a lift. Calculate the reading registered by the spring balance, in Newtons, when the lift accelerates upward at $2.5\text{ m/s}^2$ . (Take acceleration due to gravity, $g = 10\text{ m/s}^2$ ).

Show answer & explanation

Answer: 625

Answer

The reading of the spring balance is

625\text{ N}

.

When a lift accelerates upward, the effective acceleration experienced by an object relative to the lift is

g + a

. A spring balance measures the tension required to support and accelerate the object, which is

T = m(g + a)

. Substituting

m = 50\text{ kg}

,

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

, and

a = 2.5\text{ m/s}^2

gives

T = 50 \times (10 + 2.5) = 625\text{ N}

.

Step-by-Step Solution

1

Identify the given physical quantities

Mass

m = 50\text{ kg}

, acceleration due to gravity

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

, and upward acceleration

a = 2.5\text{ m/s}^2

.

Spring balances measure weight (tension or normal force), which varies in an accelerating reference frame.

2

Apply Newton's second law to determine apparent weight in an upward accelerating lift

Apparent weight (scale reading)

T = m(g + a)

.

The upward force from the spring must overcome gravitational force

mg

and provide net upward accelerating force

ma

.

3

Compute the numerical value

T = 50 \times (10 + 2.5) = 50 \times 12.5 = 625\text{ N}

.

Multiplying mass by effective acceleration yields the correct balance reading.

Key Concept

Apparent weight measurement in an accelerating frame of reference using a spring balance

Question 2Question →

A beam balance measures the mass of a metal block as $15\text{ kg}$ on Earth. The block is then transported to a planet where the acceleration due to gravity is $3.8\text{ m s}^{-2}$ and suspended from a spring balance calibrated in newtons. If the spring balance has a positive zero error of $+2.0\text{ N}$ (reading $+2.0\text{ N}$ when unloaded), what is the displayed reading on the spring balance in newtons?

Show answer & explanation

Answer: 59

Answer

The displayed reading on the spring balance is

59\text{ N}

.

The beam balance establishes that the block has a constant mass of

15\text{ kg}

. On the planet, the gravitational pull on this mass is

W = 15\text{ kg} \times 3.8\text{ m s}^{-2} = 57\text{ N}

. Because the spring balance reads

+2.0\text{ N}

when unloaded, suspending the block causes the pointer to register

57\text{ N} + 2.0\text{ N} = 59\text{ N}

.

Step-by-Step Solution

1

Determine the true mass using beam balance principles

Mass

m = 15\text{ kg}

An equal-arm beam balance measures invariant scalar mass independently of local gravitational field strength.

2

Calculate the true weight on the planet

Weight

W = 57\text{ N}

Weight is the force of gravity acting on mass, calculated as

W = mg = 15\text{ kg} \times 3.8\text{ m s}^{-2} = 57\text{ N}

.

3

Incorporate the positive zero error to find the scale pointer reading

Displayed reading =

59\text{ N}

A positive zero error means the scale indicates

+2.0\text{ N}

when no load is attached. Therefore, Scale Reading = Actual Weight + Zero Offset =

57\text{ N} + 2.0\text{ N} = 59\text{ N}

.

Key Concept

Mass is an intrinsic property measured by a beam balance, whereas weight is a force measured by a spring balance and affected by zero errors.

Estimated Time:1m 0s

Question 3Question →

An object has a mass of $12\text{ kg}$ on Earth, where the acceleration due to gravity is $10\text{ m s}^{-2}$ . What are the mass and weight of the object on the Moon, where the acceleration due to gravity is $1.6\text{ m s}^{-2}$ ?

12\text{ kg}

and

19.2\text{ N}

1.92\text{ kg}

and

19.2\text{ N}

12\text{ kg}

and

120\text{ N}

1.92\text{ kg}

and

120\text{ N}

Show answer & explanation

Answer: $12\text{ kg}$ and $19.2\text{ N}$

Answer

The mass on the Moon is

12\text{ kg}

and the weight on the Moon is

19.2\text{ N}

.

Mass is constant everywhere in the universe, so the object maintains a mass of

12\text{ kg}

on the Moon. Weight is the gravitational force exerted on the object, given by

W = mg

. On the Moon,

W = 12 \cdot 1.6 = 19.2\text{ N}

. Thus, the option specifying

12\text{ kg}

and

19.2\text{ N}

is correct.

Step-by-Step Solution

1

Determine the mass of the object on the Moon.

Mass

= 12\text{ kg}

.

Mass is a scalar physical quantity representing the quantity of matter in a body; it is constant and independent of location or gravitational field.

2

Calculate the weight of the object on the Moon using

W = m \cdot g_{\text{moon}}

.

W = 12\text{ kg} \times 1.6\text{ m s}^{-2} = 19.2\text{ N}

.

Weight is the force of gravitational attraction acting on a mass and varies directly with local gravitational field strength.

Key Concept

Mass is an intrinsic property that remains constant across different locations, whereas weight is a force that depends on local gravitational acceleration (

W = mg

).

Estimated Time:45s

Question 4Question →

A spring balance calibrated in newtons has a zero error of $+2.0\text{ N}$ (it displays $+2.0\text{ N}$ before any load is attached). When an object is suspended from the balance in a location where the acceleration due to gravity is $10\text{ m s}^{-2}$ , the scale displays a reading of $42.0\text{ N}$ . What is the true mass of the object?

4.2\text{ kg}

4.0\text{ kg}

4.4\text{ kg}

40.0\text{ kg}

Show answer & explanation

Answer: $4.0\text{ kg}$

Answer

4.0\text{ kg}

To obtain the true weight from a spring balance with a positive zero error, the zero offset must be subtracted from the indicated reading:

\text{True Weight} = 42.0\text{ N} - 2.0\text{ N} = 40.0\text{ N}

. Dividing this true weight by the gravitational acceleration (

10\text{ m s}^{-2}

) gives the true mass of

4.0\text{ kg}

.

Step-by-Step Solution

1

Correct the scale reading for instrument zero error to obtain true weight

True Weight

W = 42.0\text{ N} - 2.0\text{ N} = 40.0\text{ N}

A positive zero error means the balance overcounts force, so the initial reading must be subtracted from the scale display.

2

Calculate the true mass using the formula

W = mg

Mass

m = \frac{W}{g} = \frac{40.0\text{ N}}{10\text{ m s}^{-2}} = 4.0\text{ kg}

Mass is found by dividing the true force of gravity (weight) by the acceleration due to gravity.

Key Concept

Instrument zero error correction and mass-weight relationship

Estimated Time:1m 0s

Question 5Question →

A body is suspended from a spring balance inside an elevator that is accelerating upwards at $2.0\text{ m s}^{-2}$ on Earth, giving a scale reading of $60\text{ N}$ . The body is then taken to a planet where the acceleration due to gravity is $6.0\text{ m s}^{-2}$ and placed on an ideal beam balance against standard masses calibrated on Earth. Taking $g = 10.0\text{ m s}^{-2}$ on Earth, what is the reading obtained on the beam balance?

5.0\text{ kg}

6.0\text{ kg}

7.5\text{ kg}

3.0\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The reading obtained on the beam balance is

5.0\text{ kg}

.

In an upward accelerating elevator, the spring balance registers an apparent weight of

W_{app} = m(g + a)

. Substituting the given values gives

60\text{ N} = m(10\text{ m s}^{-2} + 2\text{ m s}^{-2})

, which yields a true mass

m = 5.0\text{ kg}

. When measured on another planet using an ideal beam balance, the gravitational force on the unknown mass balances against standard masses (

m \cdot g_p = m_{standard} \cdot g_p

). Since local gravity

g_p

cancels from both sides, the beam balance measures the invariant mass of

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the true mass of the body using the scale reading in the accelerating elevator.

m = 5.0\text{ kg}

Inside an upward accelerating elevator, the apparent weight registered by the spring balance is

W_{app} = m(g + a)

. Substituting

W_{app} = 60\text{ N}

,

g = 10.0\text{ m s}^{-2}

, and

a = 2.0\text{ m s}^{-2}

yields

60 = m(10 + 2)

, which gives

m = 5.0\text{ kg}

.

2

Determine the mass reading on a beam balance on the new planet.

The beam balance reads

5.0\text{ kg}

.

A beam balance compares the gravitational force on the object with standard masses:

m \cdot g_p = m_{standard} \cdot g_p \implies m_{standard} = m

. Because local acceleration due to gravity

g_p

cancels out on both sides, a beam balance measures true invariant mass regardless of location or gravity.

Key Concept

Mass vs Weight Measurement: Accelerating Frames and Beam Balance Invariance

Estimated Time:1m 0s

Question 6Question →

An unknown object is evaluated inside a space probe descending towards the surface of a distant planet. A beam balance calibrated on Earth registers a reading of $15.0\text{ kg}$ for the object. The probe has a downward acceleration of $2.0\text{ m s}^{-2}$ in a region where the local gravitational acceleration of the planet is $6.0\text{ m s}^{-2}$ . What reading will a spring balance display when the same object is suspended from it inside the probe?

90.0\text{ N}

120.0\text{ N}

60.0\text{ N}

150.0\text{ N}

Show answer & explanation

Answer: $60.0\text{ N}$

Answer

60.0\text{ N}

A beam balance compares gravitational forces on equal balance arms, meaning local gravity and frame acceleration affect both sides equally. Therefore, the beam balance measures the true invariant mass of

15.0\text{ kg}

. A spring balance measures the tension or apparent weight

W_{\text{app}} = m(g - a)

. Substituting

m = 15.0\text{ kg}

, local gravity

g = 6.0\text{ m s}^{-2}

, and downward acceleration

a = 2.0\text{ m s}^{-2}

gives

W_{\text{app}} = 15.0 \times (6.0 - 2.0) = 60.0\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the object using the beam balance measurement

True mass

m = 15.0\text{ kg}

A beam balance compares unknown mass with standard masses under the same local gravitational field. Because local gravity acts equally on both pans, beam balance readings yield the true invariant mass regardless of local gravitational acceleration or uniform reference frame acceleration (provided net effective gravity is greater than zero).

2

Calculate the effective acceleration experienced inside the accelerating reference frame

g_{\text{eff}} = g_{\text{planet}} - a = 6.0\text{ m s}^{-2} - 2.0\text{ m s}^{-2} = 4.0\text{ m s}^{-2}

When a reference frame accelerates downward at rate

a

, the effective apparent gravitational acceleration felt by objects suspended inside is reduced by

a

.

3

Calculate the apparent weight registered by the spring balance

W_{\text{app}} = m \cdot g_{\text{eff}} = 15.0\text{ kg} \times 4.0\text{ m s}^{-2} = 60.0\text{ N}

A spring balance measures tension force (apparent weight) exerted on its spring, which equals

m(g - a)

in a downward accelerating system.

Key Concept

Mass measurement via beam balance vs. apparent weight measurement via spring balance in accelerating frames

Question 7Question →

A spring balance and an equal-arm beam balance are used to measure an object at a location where the local acceleration due to gravity is $8.0\text{ m s}^{-2}$ . If the spring balance registers a weight of $40\text{ N}$ , what mass will the equal-arm beam balance register at this location?

5.0\text{ kg}

4.0\text{ kg}

320\text{ kg}

50\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The equal-arm beam balance will register a mass of

5.0\text{ kg}

.

Weight is given by

W = mg

. Given

W = 40\text{ N}

and

g = 8.0\text{ m s}^{-2}

, the mass of the object is

m = \frac{40}{8.0} = 5.0\text{ kg}

. An equal-arm beam balance balances the unknown mass against standard masses under the exact same local gravitational field, so local gravity cancels out and it accurately measures the object's mass as

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the true mass of the object from the spring balance reading

m = \frac{W}{g} = \frac{40\text{ N}}{8.0\text{ m s}^{-2}} = 5.0\text{ kg}

Weight is the gravitational force acting on a mass (

W = mg

), so mass equals weight divided by local acceleration due to gravity.

2

Determine the reading on the equal-arm beam balance

Mass registered =

5.0\text{ kg}

An equal-arm beam balance compares the gravitational force on the unknown mass against standard masses. Because local gravity affects both sides equally, it measures true mass independent of local gravitational acceleration.

Key Concept

Measurement of Mass and Weight

Estimated Time:1m 0s

Question 8Question →

An object of unknown mass is suspended from a spring balance possessing a zero error of $+1.5\text{ N}$ inside a lift on an unexplored planet. When the lift accelerates vertically upwards at $2.0\text{ m s}^{-2}$ , the spring balance displays a reading of $33.5\text{ N}$ . Simultaneously, an equal-arm beam balance calibrated with standard masses measures the mass of the object inside the accelerating lift to be $4.0\text{ kg}$ . What is the local acceleration due to gravity on this planet and the true weight of the object when at rest on its surface?

6.0\text{ m s}^{-2}

and

24.0\text{ N}

6.4\text{ m s}^{-2}

and

25.5\text{ N}

10.0\text{ m s}^{-2}

and

40.0\text{ N}

8.0\text{ m s}^{-2}

and

32.0\text{ N}

Show answer & explanation

Answer: $6.0\text{ m s}^{-2}$ and $24.0\text{ N}$

Answer

The local acceleration due to gravity on the planet is

6.0\text{ m s}^{-2}

and the true weight of the object at rest is

24.0\text{ N}

.

The correct response identifies that an equal-arm beam balance measures invariant mass (

4.0\text{ kg}

) because acceleration affects both balance pans equally. Subtracting the

+1.5\text{ N}

zero error from the scale reading yields a true apparent weight of

32.0\text{ N}

. Applying Newton's second law in an upward accelerating lift gives

W_{\text{app}} = m(g + a)

, which yields

32.0 = 4.0(g + 2.0)

, resulting in

g = 6.0\text{ m s}^{-2}

. The true weight at rest is therefore

W = mg = 4.0 \times 6.0 = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the true mass of the object from the beam balance measurement.

m = 4.0\text{ kg}

An equal-arm beam balance operates by comparing gravitational moments on standard masses and the test object. Because the effective acceleration

(g + a)

acts equally on both pans, it cancels out, making the beam balance measure the true, invariant mass regardless of frame acceleration or location.

2

Correct the spring balance scale reading for zero error to find the true apparent weight.

W_{\text{app}} = 33.5\text{ N} - 1.5\text{ N} = 32.0\text{ N}

A positive zero error means the balance reads

+1.5\text{ N}

when unloaded, so the true force exerted on the spring is the scale reading minus the zero error.

3

Relate apparent weight to local gravity

g

in an upward accelerating lift.

g = 6.0\text{ m s}^{-2}

In an upward accelerating frame with acceleration

a = 2.0\text{ m s}^{-2}

, the normal force/apparent weight is

W_{\text{app}} = m(g + a)

. Substituting values gives

32.0 = 4.0(g + 2.0) \implies 8.0 = g + 2.0 \implies g = 6.0\text{ m s}^{-2}

.

4

Calculate the true weight of the object when at rest on the planet.

W_{\text{true}} = 24.0\text{ N}

True weight is the force of gravity acting on the mass at rest:

W_{\text{true}} = m \cdot g = 4.0\text{ kg} \times 6.0\text{ m s}^{-2} = 24.0\text{ N}

.

Key Concept

Distinction between mass (measured by beam balance, frame-invariant) and weight (measured by spring balance, dependent on frame acceleration and zero error).

Estimated Time:2m 0s

Question 9Question →

A spring balance is attached to the ceiling of an elevator accelerating downwards at $2.0\text{ m s}^{-2}$ . Suspended from the hook of the spring balance is a light, frictionless pulley carrying two masses of $3.0\text{ kg}$ and $1.0\text{ kg}$ connected by a light inextensible string. Taking the acceleration due to gravity $g = 10.0\text{ m s}^{-2}$ , what is the reading registered by the spring balance in newtons?

Show answer & explanation

Answer: 24

Answer

The reading registered by the spring balance is 24 N.

In a frame accelerating downwards at

a = 2.0\text{ m s}^{-2}

, the effective acceleration due to gravity is reduced to

g' = g - a = 8.0\text{ m s}^{-2}

. Within this frame, the tension in the Atwood machine string is

T = \frac{2(3.0)(1.0)}{3.0 + 1.0} \times 8.0 = 12.0\text{ N}

. Since two string segments act downward on the light pulley suspended from the spring balance, the total tension force registered by the balance scale is

2T = 24.0\text{ N}

.

Step-by-Step Solution

1

Determine the effective local acceleration due to gravity inside the accelerating elevator

g' = 8.0\text{ m s}^{-2}

Because the elevator accelerates downward at

a = 2.0\text{ m s}^{-2}

, objects inside experience an apparent gravitational acceleration of

g' = g - a

.

2

Compute the tension in the string supporting the two masses in the modified gravitational field

T = 12.0\text{ N}

For an Atwood machine system in effective gravity

g'

, string tension is

T = \frac{2 m_1 m_2}{m_1 + m_2} g' = \frac{2(3.0)(1.0)}{4.0} \times 8.0 = 12.0\text{ N}

.

3

Calculate the downward pull on the spring balance

F = 24.0\text{ N}

The spring balance supports the frictionless pulley, which experiences a downward force from two upward string segments, making the total measured weight force equal to

2T = 24.0\text{ N}

.

Key Concept

Apparent weight measurement and tension forces in accelerating frames

Question 10Question →

A rocket carries a payload of mass $25\text{ kg}$ resting on a spring balance. During its vertical ascent, the spring balance indicates a reading of $350\text{ N}$ . Taking the acceleration due to gravity as $g = 10\text{ m s}^{-2}$ , what is the magnitude of the upward acceleration of the rocket in $\text{m s}^{-2}$ ?

Show answer & explanation

Answer: 4

Answer

The upward acceleration of the rocket is

4.0\text{ m s}^{-2}

.

When an object of mass

m

accelerates upward at rate

a

, the scale must exert an upward force

R

that overcomes the weight

mg

and provides net acceleration

ma

, such that

R = m(g + a)

. Substituting

R = 350\text{ N}

,

m = 25\text{ kg}

, and

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

yields

350 = 25(10 + a)

, which simplifies to

a = 4.0\text{ m s}^{-2}

.

Step-by-Step Solution

1

Formulate the equation of motion for apparent weight during upward acceleration

R - mg = ma \implies R = m(g + a)

The spring balance supports the mass against gravity while simultaneously providing the force for upward acceleration.

2

Substitute the given numerical parameters into the relation

350 = 25(10 + a)

The measured reading

R = 350\text{ N}

, mass

m = 25\text{ kg}

, and standard gravity

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

are supplied.

3

Solve the algebraic equation for the rocket's acceleration

a

a = 4.0\text{ m s}^{-2}

Dividing

350\text{ N}

by

25\text{ kg}

gives an effective total acceleration of

14\text{ m s}^{-2}

; subtracting

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

isolates the upward acceleration

a

.

Key Concept

Apparent Weight and Mass Measurement in Accelerating Frames

Question 11Question →

A spring balance calibrated in newtons shows an initial pointer reading of $+0.4\text{ N}$ before any load is suspended. When a stone is attached to the balance hook, the pointer indicates $18.4\text{ N}$ . Taking acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the actual mass of the stone?

1.8\text{ kg}

1.84\text{ kg}

1.88\text{ kg}

18.0\text{ kg}

Show answer & explanation

Answer: $1.8\text{ kg}$

Answer

The actual mass of the stone is

1.8\text{ kg}

.

The spring balance has a systematic positive zero error of

+0.4\text{ N}

. To find the true weight exerted by the stone, this offset must be subtracted from the observed scale reading:

18.4\text{ N} - 0.4\text{ N} = 18.0\text{ N}

. Dividing this true weight by the acceleration due to gravity (

10\text{ m s}^{-2}

) yields the correct mass of

1.8\text{ kg}

.

Step-by-Step Solution

1

Determine the true weight by applying zero error correction

W_{\text{true}} = 18.4\text{ N} - 0.4\text{ N} = 18.0\text{ N}

A positive zero error means the scale overstates the load and must be subtracted from the observed reading.

2

Calculate mass using the relation between weight, mass, and gravitational field strength

m = \frac{W_{\text{true}}}{g} = \frac{18.0\text{ N}}{10\text{ m s}^{-2}} = 1.8\text{ kg}

Weight is equal to mass multiplied by acceleration due to gravity (

W = mg

).

Key Concept

Measurement of Mass and Weight with Zero Error Correction

Question 12Question →

A chemical balance is used to determine the mass of a metallic sphere, giving a reading of $6.0\text{ kg}$ . The sphere is then suspended from a spring balance at a laboratory where the local acceleration due to gravity is $10\text{ m s}^{-2}$ . If the spring balance reads $-3.0\text{ N}$ before any load is attached, what is the indicated reading on the spring balance when the sphere is suspended?

57.0\text{ N}

63.0\text{ N}

60.0\text{ N}

6.0\text{ N}

Show answer & explanation

Answer: $57.0\text{ N}$

Answer

The indicated reading on the spring balance is

57.0\text{ N}

.

The true weight of the metallic sphere is calculated as

W = mg = 6.0\text{ kg} \times 10\text{ m s}^{-2} = 60.0\text{ N}

. Because the spring balance has an initial zero reading of

-3.0\text{ N}

before any load is attached, the indicated reading when the sphere is hung is

60.0\text{ N} - 3.0\text{ N} = 57.0\text{ N}

.

Step-by-Step Solution

1

Calculate the true weight of the metallic sphere

W = mg = 6.0\text{ kg} \times 10\text{ m s}^{-2} = 60.0\text{ N}

Weight is the gravitational force acting on mass.

2

Set up the zero error relationship for instrument measurement

\text{Actual Weight} = \text{Indicated Reading} - \text{Zero Reading}

The actual physical value equals the observed pointer reading minus the zero reading of the unloaded instrument.

3

Substitute known values and solve for the indicated reading (

R

)

60.0\text{ N} = R - (-3.0\text{ N}) \implies R = 60.0 - 3.0 = 57.0\text{ N}

Solving the linear relation yields an indicated pointer position of

57.0\text{ N}

.

Key Concept

Mass versus Weight Measurement and Zero Error Correction

Estimated Time:1m 0s

Question 13Question →

A body is weighed on a distant planet where the acceleration due to gravity is $4\text{ m s}^{-2}$ . A spring balance calibrated on Earth ( $g = 10\text{ m s}^{-2}$ ) gives a reading of $20\text{ N}$ for the body on this planet. What is the mass of the body as measured by a beam balance on the planet?

2.0\text{ kg}

5.0\text{ kg}

50.0\text{ kg}

80.0\text{ kg}

Show answer & explanation

Answer: $5.0\text{ kg}$

Answer

The mass of the body as measured by a beam balance is

5.0\text{ kg}

.

The spring balance measures weight (

W = mg_{\text{planet}}

). Substituting the given values (

20\text{ N} = m \times 4\text{ m s}^{-2}

) gives

m = 5.0\text{ kg}

. Since mass is constant everywhere and a beam balance measures mass independently of local gravity variation, the beam balance reads

5.0\text{ kg}

.

Step-by-Step Solution

1

Calculate the mass of the body using the spring balance reading on the planet.

m = \frac{W_{\text{planet}}}{g_{\text{planet}}} = \frac{20\text{ N}}{4\text{ m s}^{-2}} = 5.0\text{ kg}

Spring balances measure weight force (

W = mg

). The local weight divided by local gravity yields the mass.

2

Determine the reading on a beam balance.

The beam balance measures mass directly by comparison, yielding

5.0\text{ kg}

.

Mass is an invariant property of matter and does not change with location or gravitational field strength.

Key Concept

Mass vs. Weight and Instrument Principles

Estimated Time:1m 0s

Question 14Question →

A spring balance is calibrated in kilograms at sea level where $g = 10.0\text{ m s}^{-2}$ . An object measured with an equal-arm beam balance at sea level is found to have a mass of $36.0\text{ kg}$ . If this object is taken to a high altitude where $g = 9.0\text{ m s}^{-2}$ and suspended from the same spring balance, what reading will the scale of the spring balance display?

32.4\text{ kg}

36.0\text{ kg}

40.0\text{ kg}

324.0\text{ kg}

Show answer & explanation

Answer: $32.4\text{ kg}$

Answer

The spring balance scale will display a reading of

32.4\text{ kg}

.

An equal-arm beam balance measures true invariant mass (

36.0\text{ kg}

) by comparing gravitational moments on two arms. A spring balance measures force (weight). At high altitude, the gravitational force acting on the mass is

W = 36.0\text{ kg} \times 9.0\text{ m s}^{-2} = 324\text{ N}

. Because the spring balance scale was calibrated assuming sea-level gravity (

10.0\text{ m s}^{-2}

), it converts force to indicated mass as

324\text{ N} / 10.0\text{ m s}^{-2} = 32.4\text{ kg}

.

Step-by-Step Solution

1

Calculate the true weight of the object at the new altitude

W = m \times g_{\text{altitude}} = 36.0\text{ kg} \times 9.0\text{ m s}^{-2} = 324\text{ N}

Weight is the force of gravity acting on a mass at a specific location.

2

Determine the indicated mass reading on the calibrated spring balance scale

\text{Reading} = \frac{W}{g_{\text{calibration}}} = \frac{324\text{ N}}{10.0\text{ m s}^{-2}} = 32.4\text{ kg}

The spring balance measures tension force but its scale was calibrated using sea-level gravity (

10.0\text{ m s}^{-2}

) to indicate mass.

Key Concept

Mass is an intrinsic property measured independently of gravity by a beam balance, whereas weight is a force measured by a spring balance, causing mass-calibrated spring balances to give different readings under varying local gravitational accelerations.

Estimated Time:1m 0s

Question 15Question →

A spring balance suspended inside an elevator reads $48\text{ N}$ when supporting an object while the elevator accelerates downwards at $2\text{ m s}^{-2}$ . Taking acceleration due to gravity $g = 10\text{ m s}^{-2}$ , what is the mass of the object as measured by an equal-arm beam balance?

6.0\text{ kg}

4.0\text{ kg}

4.8\text{ kg}

60.0\text{ kg}

Show answer & explanation

Answer: $6.0\text{ kg}$

Answer

The mass of the object measured by an equal-arm beam balance is

6.0\text{ kg}

.

In a downward accelerating elevator, the apparent weight indicated by a spring balance is

W_{app} = m(g - a)

. Substituting

48\text{ N} = m(10\text{ m s}^{-2} - 2\text{ m s}^{-2}) = 8m

gives a mass of

6.0\text{ kg}

. Because a beam balance compares masses directly and both sides experience the exact same effective acceleration, it measures true mass regardless of the frame's acceleration.

Step-by-Step Solution

1

Formulate the equation for apparent weight in a downward accelerating reference frame.

W_{app} = m(g - a)

When an elevator accelerates downward, the effective acceleration experienced by an object inside is

(g - a)

.

2

Substitute the given values (

W_{app} = 48\text{ N}

,

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

,

a = 2\text{ m s}^{-2}

) into the formula to find mass

m

.

48 = m(10 - 2) \implies 48 = 8m \implies m = 6.0\text{ kg}

This yields the true scalar mass of the object.

3

Determine the reading on an equal-arm beam balance.

The beam balance reads

6.0\text{ kg}

.

A beam balance compares unknown mass against standard counter-weights; both experience identical effective acceleration, making its measurement independent of motion or local gravity.

Key Concept

Apparent weight in an accelerating frame vs invariant mass measurement using a beam balance

Estimated Time:1m 0s

Question 16Question →

Match each physical quantity or measuring instrument related to mass and weight in Column A with its corresponding physical property or operational principle in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Beam balance matches with comparing gravitational forces using the principle of moments; Spring balance matches with measuring gravitational pull directly based on Hooke's law; Mass matches with an intrinsic scalar quantity measured in kilograms; Weight matches with a downward vector quantity measured in newtons.

Beam balances measure mass via the principle of moments independently of gravitational variation. Spring balances measure weight force using spring extension under Hooke's law. Mass is a constant scalar quantity measured in kilograms, while weight is a variable vector force measured in newtons.

Step-by-Step Solution

1

Identify the working principles of mass and weight measuring instruments.

The beam balance uses equal arms to balance moments (

m_1 g L = m_2 g L

), canceling

g

, so it measures invariant mass. The spring balance measures the force pulling a spring (

F = kx = mg

), depending directly on local

g

.

Instrument operation dictates whether mass or weight is measured.

2

Distinguish between the physical properties of mass and weight.

Mass is a scalar measure of quantity of matter (SI unit:

\text{kg}

) and is constant. Weight is the gravitational force acting on mass (SI unit:

\text{N}

) and is a vector quantity.

Fundamental physical definitions set the units and vector/scalar characteristics of mass versus weight.

Key Concept

Operating principles of balances and fundamental properties distinguishing mass from weight.

Estimated Time:1m 0s

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