Mass — Question 1

Question 1

Does moving the object change its mass?

An unchanged object of mass $2.50\ \mathrm{kg}$ is moved between two stationary laboratories. Treat the data below as exact in the model; ignore air buoyancy.

For this exercise the gravitational force is $W=mg$. A force sensor displays $W$ in newtons. A second instrument measures that force but displays a number in kilograms by dividing by the fixed calibration value $g_0=9.80\ \mathrm{N\,kg^{-1}}$.

An ideal equal-arm balance is also available. It compares the object with reference masses in the same local gravitational field; it is level when the two gravitational forces are equal.

1. Find the force sensor reading and the force-based kilogram display in each laboratory.

2. What total reference mass balances the object in each laboratory? Explain which factor cancels in the comparison and why it does not cancel in the unadjusted kilogram display.

3. A student interprets the smaller kilogram display in B as loss of material. Assess that conclusion and state a correction that would make this display report mass correctly in B.

4. Could either gravitational weighing method determine the object’s mass in an ideal region where $g=0$? Does the loss of weighing sensitivity imply zero mass?

Question 1 - Solution

Does moving the object change its mass?

(a) Separate the force from the displayed mass estimate.

In A, $W_A=(2.50)(9.80)\ \mathrm{N}=\boxed{24.5\ \mathrm{N}}$, so the mass display reads $24.5/9.80=\boxed{2.50\ \mathrm{kg}}$. In B, $$W_B=(2.50)(1.60)\ \mathrm{N}=\boxed{4.00\ \mathrm{N}},\qquad m_{\mathrm{disp,B}}=\frac{4.00\ \mathrm{N}}{9.80\ \mathrm{N\,kg^{-1}}} \approx\boxed{0.40816\ \mathrm{kg}}.$$

(b) A local comparison cancels a common nonzero field.

$$mg=m_{\mathrm{ref}}g\ \overset{g\ne0}{\Longrightarrow}\ m=m_{\mathrm{ref}},\qquad m_{\mathrm{disp}}=m\frac g{g_0}.$$

At balance, $mg=m_{\mathrm{ref}}g$. For either stated laboratory, division by the same nonzero $g$ gives $\boxed{m_{\mathrm{ref}}=2.50\ \mathrm{kg}}$. The force-based display instead gives $m_{\mathrm{disp}}=mg/g_0$. Only when $g=g_0$ does it equal the object’s mass without adjustment.

(c) A changed force does not establish a changed mass.

$$m_{\mathrm{corrected,B}}=m_{\mathrm{disp,B}}\frac{g_0}{g_B}=\frac{4.00\ \mathrm N}{1.60\ \mathrm{N\,kg^{-1}}}=2.50\ \mathrm{kg}.$$

The prompt stipulates that the object is unchanged. The low display in B is a calibration mismatch: the instrument divides a force measured in B by the value of $g$ for A. Divide by $1.60\ \mathrm{N\,kg^{-1}}$ instead, or calibrate the mass display against known masses in B. The result is 2.50 kg.

(d) Zero gravitational signal is not zero mass.

When $g=0$, every mass gives $W=0$. The balance condition becomes $0=0$ and distinguishes no masses; the force sensor also supplies no mass information. The object still has mass. A method based on its response to a controlled nongravitational interaction would be needed.