Question

In Figure, a chain is held on a frictionless table with one fourth of its length hanging over the edge. If the chain has lengthL = 28 cmand mass m = 0.012 kg, how much work is required to pull the hanging part back onto the table?

Short answer

The work required to pull the hanging part back onto the table is$W=0.0010\mathrm{J}$

Step-by-step solution

Given

1. The hanging length of the chain is,$I=L/4$
2. The length of the chain is,$L=28\mathrm{cm}=0.28\mathrm{m}$
3. The mass of the chain is, $m=0.012\mathrm{kg}$

Determining the concept

The problem deals with the law of conservation of energy. According to the law of energy conservation, energy can neither be created, nor be destroyed.Use the law of conservation of energy andtheformula for the work done in terms of force and small displacement. Infinitesimally small work done can be integrated over a given range of displacement to find the total work done.

Formulae are as follow:

$$\begin{gathered} U=mgh \\ W={\int }_{xi}^{xf}Fdx \end{gathered}$$

where, Uis potential energy, m is mass, g is an acceleration due to gravity, x is displacement, his height, F is force and W is work done.

Determining thework required to pull the hanging part back onto the table

Work is required to pull the hanging part back onto the table. This work done is the change of gravitational potential energy. The changing chain can be divided into a large number of infinitesimal segments. Consider, the length of the segment is $dy$and the mass of the segment is $\left(\frac{m}{L}\right)dy$. The vertical distance covered by the chain is $-y$onto the table.

The chain potential energy of the segment is,

$$\begin{gathered} dU=mgh \\ dU=\left(\frac{m}{L}\right)gydy \end{gathered}$$

During the motion of the chain, it moves from the lower end to the origin at the table top. The total potential energy is work done by the variable force,

role="math" localid="1661223837351" $$\begin{gathered} W={\int }_{xi}^{xf}Fdx \\ W=∆U={\int }_{\left(-U4\right)}^0-\left(\frac{m}{L}\right)gydy \\ W=∆U=-\left(\frac{m}{L}\right)g{\int }_{\left(-U4\right)}^{0}ydy \\ W=∆U=-\left(\frac{m}{L}\right)g{\left[\frac{y^2}{2}\right]}_{\left(-U4\right)}^0 \\ W=∆U=\frac{1}{2}\left(\frac{m}{L}\right)g{\left(\frac{L}{4}\right)}^2 \\ W=∆U=\frac{mgL}{32} \\ W=\frac{0.012\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.28\mathrm{m}}{32} \\ W=0.0010\mathrm{J} \end{gathered}$$

Hence, the work required to pull the hanging part back onto the table is $W=0.0010\mathrm{J}$

Therefore, the work required to pull the hanging part back onto the table by using the concept of gravitational potential energy and work done by variable force.