Question

A uniform chain of length \(2 \mathrm{~m}\) is kept on a table such that a length of \(50 \mathrm{~cm}\) hangs freely from the edge of the table. The total mass of the chain is \(5 \mathrm{~kg}\). What is the work done in pulling the entire chain on the table. \(\left(\mathrm{g}=10 \mathrm{~m} \mathrm{~s}^{2}\right)\) (A) \(7.2 \mathrm{~J}\) (B) \(3 \mathrm{~J}\) (C) \(4.6 \mathrm{~J}\) (D) \(120 \mathrm{~J}\)

Short answer

The work done in pulling the entire chain on the table is approximately \(3 \mathrm{~J}\).

Step-by-step solution

Convert length to meters

The length of the chain that is dangling from the table is given as 50 cm. We know that 1m = 100cm, so we convert 50 cm to meters, which gives us 0.5 meters.

Find the mass per unit length of the chain

This can be calculated by dividing the total mass by the total length of the chain. The total mass is 5 kg, and the total length is 2 m. So, \[ \text{Mass per unit length} = \frac{5 kg}{2 m} = 2.5 \, \frac{kg}{m} \]

Calculate the weight of the dangling portion of the chain at the start

This is equal to the mass per unit length multiplied by the length of the chain dangling from the table, and further multiplied by g (acceleration due to gravity), assuming downward direction as negative. So, we get \[ \text{Weight} = -2.5 \frac{kg}{m} \times 0.5m \times 10 \frac{m}{s^2} = -12.5 \, \text{N} \]

Find the work done

Now we find the work done by integrating the varying force (weight) with respect to the distance it acts on. We have to integrate from 0 to 0.5 (length of the dangling part) because the force decreases from -12.5 N (at start) to 0 N (when fully pulled up). The force function in this case is directly proportional to the length of the chain hanging from the table. The integral can be calculated as follow: \[ \text{Work Done} = \int_0^{0.5} (-12.5) \frac{y}{0.5} \, dy = -12.5 \int_0^{0.5} 2y \, dy \] This integration gives -12.5 * [y^2] from 0 to 0.5. Substituting these limits, the resulting work done is \[ \text{Work Done} = -12.5*[ (0.5)^2 - (0)^2] = -12.5 * 0.25 = -3.125 \, \text{J} \] However, work done is a scalar quantity and shouldn't be negative. The negative sign here just indicates that the work was done against the weight of the chain. So, the magnitude of work done will be: \[ \text{Work Done} = 3.125 \, \text{J} \] So, the closest answer is (B) \(3 \mathrm{~J}\).

Key concepts

Uniform Chain Problem

When dealing with the Uniform Chain Problem, it involves analyzing how a chain behaves when a part of it is hanging freely under the influence of gravity. This uniform chain is typically extended over an edge of a table or some surface.

The challenge usually involves determining specific physical properties like gravitational force effects, work done in pulling the chain back up, or energy transformations. Understanding this requires:

• Considering the chain's uniformity, which means its mass per unit length is constant.
• Examining how much of the chain's length is hanging and how it affects the total weight of the dangling section.
• Calculating the forces involved and any required work or energy changes to move or lift the chain.

Solving these problems allows us to illustrate important principles in mechanics, like how energy is required to perform work against gravitational forces.

Gravitational Potential Energy

Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field. It is directly related to the height at which the object is located and the object's mass. In the context of the Uniform Chain Problem, GPE is critical as it involves the chain segment that is hanging from the table.

The formula used for GPE is: \[U = mgh\]Where:

• \( m \) is the mass of the object,
• \( g \) is the acceleration due to gravity (usually \( 9.8 \, \text{m/s}^2 \), but can be approximated as 10 \( \, \text{m/s}^2 \)).
• \( h \) is the height above the reference point.

In the problem at hand, the center of mass of the dangling part of the chain is considered to be at the midpoint of that portion. This is because gravitational potential energy is calculated based on where the entire mass of that part of the chain seems to "act from".

As you pull the chain back onto the table, you are effectively moving this center of mass upward, increasing its GPE.

Work Done by Gravity

The concept of Work Done by Gravity is crucial in understanding the energy changes when lifting objects against gravitational forces. When a portion of the chain hangs off the table and is pulled up, work must be done against gravitational pull.

Work done in physics is calculated using: \[W = \int F \cdot dx\]For the chain's scenario, the force \( F \) is the weight of the chain's hanging portion. This weight varies as portions of the chain are lifted back onto the table. Initially, the entire overhanging segment exerts full weight, decreasing linearly to zero as it is fully pulled up.

The work you do to pull the chain up is equivalent to:

• Overcoming the gravitational pull on each infinitesimally small section of chain.
• Changing the gravitational potential energy as the height of the center of mass increases.

This work done is a direct result of energy transfer – from the mechanical energy you exert to the increase in gravitational potential energy of the chain. Work is done when there is movement, and in this problem, it's the movement of the chain to a higher position on the table.