An object of unknown mass is suspended from a spring balance possessing a zero error of +1… — JAMB UTME

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 $24.0\text{ N}$ Answer
B
$6.4\text{ m s}^{-2}$ and $25.5\text{ N}$
C
$10.0\text{ m s}^{-2}$ and $40.0\text{ N}$
D
$8.0\text{ m s}^{-2}$ and $32.0\text{ N}$

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

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.

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.

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}

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

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