© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Problem
A uniform cable of mass $M$ and length $L$ is placed on a horizontal surface such that its $(1/n)$th part hangs off the edge. We need to find the work required to lift the hanging portion (length $L/n$) back onto the surface.
Step 2: Determine the Linear Mass Density
The total mass of the cable is $M$, and its total length is $L$. Therefore, its linear mass density (mass per unit length) is
$$\lambda = \frac{M}{L}.$$
Step 3: Set Up an Elemental Section
Consider a small element of the cable at a distance $x$ from the edge of the table (measured downward into the hanging part). The thickness of this small element is $dx$. The mass of this small element is
$$dm = \lambda \, dx = \frac{M}{L}\,dx.$$
Step 4: Calculate the Small Amount of Work $dW$
The weight of this small element is $dm \, g = \frac{M}{L} g \, dx$. This element must be lifted a distance $x$ to reach the surface. Hence, the small work needed to lift this element is:
$$dW = \text{(weight of element)} \times \text{(distance lifted)} = \left(\frac{M}{L} g\,dx\right)\times x = \frac{M\,g}{L} x \, dx.$$
Step 5: Integrate from $x = 0$ to $x = \frac{L}{n}$
To find the total work $W$, sum (integrate) $dW$ over the entire length of the hanging portion, which goes from $x=0$ (where the cable meets the table) to $x=\frac{L}{n}$ (the free hanging end):
$$
W = \int_{0}^{L/n} \frac{M\,g}{L}\,x \, dx.
$$
Step 6: Compute the Integral
Evaluate the integral:
\[
W = \frac{M\,g}{L}\,\int_{0}^{L/n} x \,dx
= \frac{M\,g}{L} \left[\frac{x^2}{2}\right]_{0}^{L/n}
= \frac{M\,g}{L} \times \frac{\left(\frac{L}{n}\right)^2}{2}
= \frac{M\,g}{L} \times \frac{L^2}{2n^2}
= \frac{M\,g\,L}{2n^2}.
\]
Step 7: State the Final Answer
The total work done to lift the hanging portion of the cable back onto the surface is:
\[
\boxed{\frac{MgL}{2n^2}}.
\]
