A mass is attached to a spring that hangs vertically. The extension produced in the spring is 6 cm on Earth. The acceleration due to gravity on the surface of the Moon is one-sixth of its value on the surface of the Earth. The extension of the spring on the Moon would be:
1 cm
Understanding Spring Extension and Gravity
This problem involves a spring supporting a mass and how its extension changes when the gravitational force acting on the mass changes. The key principles are Hooke's Law and the relationship between mass, gravity, and weight.
Applying Hooke's Law on Earth
Hooke's Law states that the force applied to a spring is directly proportional to its extension, provided the elastic limit is not exceeded. Mathematically, this is expressed as:
\( F = kx \)
Where:
- \(F\) is the force applied to the spring.
- \(k\) is the spring constant, a measure of the spring's stiffness.
- \(x\) is the extension (or compression) of the spring from its natural length.
In this case, the force stretching the vertical spring is the weight of the attached mass. The weight \(W\) of an object is given by:
\( W = mg \)
Where:
- \(m\) is the mass of the object.
- \(g\) is the acceleration due to gravity.
On Earth, let the acceleration due to gravity be \(g_E\) and the extension be \(x_E\). The weight of the mass is \(W_E = mg_E\). According to Hooke's Law, this weight is balanced by the spring force:
\( mg_E = kx_E \)
We are given that the extension on Earth is 6 cm (\(x_E = 6\) cm). So, the equation becomes:
\( mg_E = k(6) \quad \text{(Equation 1)} \)
We can express the spring constant \(k\) in terms of \(m\) and \(g_E\) from this equation:
\( k = \frac{mg_E}{6} \)
Analyzing the Scenario on the Moon
Now, the same mass and the same spring are taken to the Moon. The acceleration due to gravity on the Moon (\(g_M\)) is one-sixth of its value on Earth:
\( g_M = \frac{1}{6} g_E \)
Let the extension of the spring on the Moon be \(x_M\). The weight of the mass on the Moon is \(W_M = mg_M\). Applying Hooke's Law on the Moon:
\( mg_M = kx_M \)
Substitute the value of \(g_M\) and the expression for \(k\) that we found from the Earth scenario:
\( m \left(\frac{1}{6} g_E\right) = \left(\frac{mg_E}{6}\right) x_M \)
Calculating the Extension on the Moon
Now, we need to solve this equation for \(x_M\):
\( \frac{1}{6} mg_E = \frac{mg_E}{6} x_M \)
Notice that the term \( \frac{mg_E}{6} \) appears on both sides of the equation. We can cancel this term out (assuming \(m \neq 0\) and \(g_E \neq 0\)):
\( 1 = x_M \)
Therefore, the extension of the spring on the Moon would be 1 cm.
Summary of Results
| Parameter | On Earth | On Moon |
|---|---|---|
| Mass (\(m\)) | Same | Same |
| Spring Constant (\(k\)) | Same | Same |
| Gravity (\(g\)) | \(g_E\) | \(g_M = \frac{1}{6} g_E\) |
| Weight (\(W = mg\)) | \(mg_E\) | \(mg_M = m(\frac{1}{6} g_E) = \frac{1}{6} mg_E\) |
| Extension (\(x\)) | \(x_E = 6\) cm | \(x_M\) |
| Hooke's Law (\(mg=kx\)) | \(mg_E = k(6)\) | \(mg_M = kx_M \implies \frac{1}{6} mg_E = kx_M\) |
Since \(mg_E = 6k\), we can substitute \(6k\) for \(mg_E\) in the Moon equation:
\( \frac{1}{6} (6k) = kx_M \)
\( k = kx_M \)
Assuming \(k \neq 0\) (which it must be for the spring to extend), we get:
\( 1 = x_M \)
The extension on the Moon is 1 cm.
Revision Table: Spring Extension Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Hooke's Law | \(F = kx\). Relates force on a spring to its extension. | Used to model the relationship between the weight and spring extension. |
| Weight | \(W = mg\). Force due to gravity acting on a mass. | The force causing the spring extension is the weight of the mass, which changes with gravity. |
| Spring Constant (\(k\)) | A measure of spring stiffness; constant for a given spring. | Remains the same whether the spring is on Earth or the Moon. |
| Acceleration due to Gravity (\(g\)) | Varies depending on the celestial body (Earth, Moon, etc.). | The change in gravity from Earth to Moon is the primary reason for the change in spring extension. |
Additional Information: Hooke's Law and Weight
Hooke's Law is fundamental in understanding elastic behavior. It applies to many elastic materials, not just springs, within their elastic limit. Beyond this limit, the material does not return to its original shape, and Hooke's Law no longer holds.
Weight is a force, measured in Newtons (N) in the SI system. Mass, on the other hand, is an intrinsic property of an object and measures the amount of matter it contains, measured in kilograms (kg). Mass remains constant regardless of location, but weight changes depending on the local gravitational acceleration.
In this problem, the mass attached to the spring is the same on Earth and the Moon. The spring itself also has the same stiffness, meaning its spring constant \(k\) is unchanged. The only thing that changes is the gravitational pull, which affects the weight of the mass, and thus the force stretching the spring.
Since the weight on the Moon is one-sixth of the weight on Earth (\(W_M = \frac{1}{6} W_E\)), and the extension is directly proportional to the force (\(x \propto F\)) for a given spring (constant \(k\)), the extension on the Moon will also be one-sixth of the extension on Earth.
\( x_M = \frac{1}{6} x_E \)
\( x_M = \frac{1}{6} (6 \text{ cm}) \)
\( x_M = 1 \text{ cm} \)
This confirms the result obtained through detailed calculation.
Important Questions from Miscellaneous
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1. Distance between the longitudes becomes zero on North Pole and South Pole.
2. Distance between the longitudes is maximum on the Equator.
3. Number of longitudes is more than number of latitudes.
Which of the statements given above is/are correct?
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