To find the work done in pulling a uniform rope of linear density 'd' and length '1' onto the table, we need to consider the work required to lift each infinitesimally small piece of the rope.
Let's represent:
- The linear density of the rope as [tex]\(d\)[/tex], which is the mass per unit length.
- The length of the rope as [tex]\(L = 1\)[/tex].
- The acceleration due to gravity as [tex]\(g\)[/tex].
We are required to pull the entire rope onto the table, which means raising each small segment of the rope from its hanging position to the level of the table.
Firstly, let's denote a small segment of the rope at a distance [tex]\(x\)[/tex] from the top of the table. The weight of this small segment [tex]\(dx\)[/tex] of the rope is:
[tex]\[ d \cdot dx \][/tex]
The potential energy required to lift this segment a height [tex]\(x\)[/tex] (from its initial hanging position to the top of the table) is:
[tex]\[ d \cdot g \cdot x \cdot dx \][/tex]
Next, we need to sum (integrate) the potential energy required to lift all such segments from [tex]\(x = 0\)[/tex] to [tex]\(x = 1\)[/tex] (the entire length of the rope):
[tex]\[
\text{Total work done} = \int_{0}^{1} d \cdot g \cdot x \, dx
\][/tex]
Let's evaluate this integral step-by-step.
[tex]\[
\text{Integrating} \quad \int_{0}^{1} d \cdot g \cdot x \, dx
\][/tex]
The linear density [tex]\(d\)[/tex] and the acceleration due to gravity [tex]\(g\)[/tex] are constants, so they can be taken outside the integral:
[tex]\[
= d \cdot g \int_{0}^{1} x \, dx
\][/tex]
Now, integrate [tex]\(x \, dx\)[/tex]:
[tex]\[
\int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1}
\][/tex]
Evaluating this at the bounds 0 and 1:
[tex]\[
= \left. \frac{x^2}{2} \right|_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}
\][/tex]
So the total work done is:
[tex]\[
d \cdot g \cdot \frac{1}{2} = \frac{d g}{2}
\][/tex]
Therefore, the work done in pulling the rope onto the table is:
[tex]\[
\boxed{\frac{d g}{2}}
\][/tex]
So, the correct answer is:
1) [tex]\(\frac{d g}{2}\)[/tex]
