A body of mass 50\text{ kg} is suspended from a spring balance attached to the ceiling of … — JAMB UTME

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, $g = 10\text{ m/s}^2$ ).

Answer: 625 N

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

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.

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

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

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