A spring balance calibrated in newtons has a zero error of +2.0\text{ N} (it displays +2.0… — JAMB UTME

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?

A
$4.2\text{ kg}$
$4.0\text{ kg}$ Cevap
C
$4.4\text{ kg}$
D
$40.0\text{ kg}$

Cevap

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}

Adım Adım Çözüm

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.

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.

Anahtar Kavram

Instrument zero error correction and mass-weight relationship

Tahmini Süre:1m 0s

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