Kvant Physics Problem 183

The problem involves a dynamometer sliding along a smooth horizontal table, pulled by a constant horizontal force $F=4~\text{N}$, with the mass of its spring equal to the mass of its casing.

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Verdicts: FAIL + FAIL
Solve time: 16m00s
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Problem

A dynamometer sliding along a smooth horizontal table is pulled with a constant force $F=4~\text{Н}$. What does the dynamometer read if the mass of its spring is equal to the mass of its casing, and the dynamometer was calibrated in the horizontal position?

G. L. Kotkin

Checks

The problem involves a dynamometer sliding along a smooth horizontal table, pulled by a constant horizontal force $F=4~\text{N}$, with the mass of its spring equal to the mass of its casing. The dynamometer was calibrated in the horizontal position, so it measures the force transmitted through the spring when the system is in motion. The relevant physical principles are Newton's second law for a system of connected masses and the definition of the reading of a spring dynamometer, which is the internal force in the spring. No electromagnetic effects are involved.

Let $m_c$ denote the mass of the casing and $m_s$ the mass of the spring. According to the problem, $m_s = m_c$. When a constant horizontal force $F$ is applied to the dynamometer as a whole, the total mass being accelerated is $m_\text{total} = m_c + m_s = 2 m_c$. The acceleration of the system is therefore

$a = \frac{F}{m_\text{total}} = \frac{F}{2 m_c}.$

The spring inside the dynamometer connects the pulling point to the casing. In the accelerated frame, the reading of the dynamometer is the force transmitted through the spring to the casing, which is the force required to accelerate the casing alone at the system acceleration. The force on the casing is therefore

$F_\text{spring} = m_c , a = m_c \cdot \frac{F}{2 m_c} = \frac{F}{2}.$

Substituting the given force $F = 4~\text{N}$, the dynamometer reads

$F_\text{spring} = \frac{4~\text{N}}{2} = 2~\text{N}.$

This calculation correctly accounts for the fact that the spring has non-negligible mass. If the spring were massless, the entire applied force $F$ would be transmitted to the casing, and the dynamometer would read $F$. When the spring has mass equal to the casing, the reading is reduced by a factor of two, reflecting the division of the total force between accelerating the spring and the casing.

The units are consistent: acceleration has units $\text{m/s}^2$, force has units $\text{N} = \text{kg·m/s}^2$, and the computed dynamometer reading is in newtons. The limiting cases behave as expected. If the spring mass $m_s \to 0$, the reading approaches $F$, corresponding to a massless spring transmitting the full force. If $m_s \to \infty$, the reading approaches zero, since the spring itself absorbs the acceleration and the casing barely moves.

Summary

The solution identifies the correct physical principles, explicitly derives the system acceleration, and computes the force transmitted through the spring, which corresponds to the dynamometer reading. The algebraic steps are correct, units are consistent, and limiting cases are physically reasonable. The final answer is

$F_\text{spring} = 2~\text{N}.$

All steps are fully justified, and the mass of the spring is properly incorporated into the analysis.

VERDICT: PASS.