Problem 14 - Single Variable Calculus

Question

A chain with length \(5 \mathrm{~m}\) and mass \(30 \mathrm{~kg}\) is lying on the ground. Find the work done in pulling one end of the chain vertically upward to a height of \(2 \mathrm{~m}\).

Step-by-Step Solution

Answer

The work done in pulling one end of the chain vertically upward to a height of 2 meters is \(471.12 \mathrm{~J}\).

Step 1: Define the mass distribution of the chain

Let's define the linear mass density (mass per unit length) of the chain: \(\lambda = \frac{m}{L}\), where \(m = 30 \mathrm{~kg}\) and \(L = 5 \mathrm{~m}\). Therefore, \(\lambda = \frac{30}{5} = 6 \mathrm{~kg/m}\).

Step 2: Set up the integral

Now we need to find the work done in lifting the chain incrementally as its height changes from 0 to 2 meters. To do this, we can set up an integral for the work done as follows: \(W = \int_{0}^{2} F(h) \, dh\), where \(h\) represents the height and \(F(h)\) is the gravitational force acting on each incremental segment of the chain at height \(h\), given by: \(F(h) = \lambda g (L - h)\), where \(\lambda\) is the linear mass density, \(g = 9.81 \mathrm{~m/s^2}\) is the gravitational acceleration, and \(L\) is the total length of the chain.

Step 3: Calculate the integral

Now we can plug the values we found in step 1 into the integral to find the work done: \(W = \int_{0}^{2} 6(9.81)(5-h) \, dh = 6(9.81) \int_{0}^{2} (5-h) \, dh\). To calculate this integral, we can use the power rule: \(6(9.81) \int_{0}^{2} (5-h) \, dh = 6(9.81) [(5h - \frac{1}{2}h^2)|_{0}^{2}] \).

Step 4: Evaluate the integral and find the work done

Evaluate the integral and find the work done: \(W = 6(9.81)[(5\cdot2 - \frac{1}{2}(2^2)) - (5\cdot0 - \frac{1}{2}(0^2))] = 6(9.81)(10 - 2) = 6(9.81)(8)\). So the work done in pulling one end of the chain vertically upward to a height of 2 meters is: \(W = 6(9.81)(8) = 471.12 \mathrm{~J}\). Therefore, the work done in pulling one end of the chain vertically upward to a height of 2 meters is 471.12 Joules.

Key Concepts

Mass Distribution

In physics, mass distribution refers to how mass is spread out over a certain space, such as a chain or rod. When dealing with physical problems like a chain or a rope being lifted, understanding the mass distribution is essential.
Consider a chain lying flat on the ground: its mass is spread evenly over its entire length. If we say the chain is 5 meters long and has a mass of 30 kg, we can determine how this mass is distributed along its length.
Using the concept of linear mass density, we divide the total mass by the total length. This gives us the mass per unit length, which tells us how much mass there is in each meter of the chain. In our example, this calculation gives us a linear mass density of 6 kg/m.

Linear Mass Density

Linear mass density represents the amount of mass per unit length of an object, like a rope or chain. It is a crucial concept when dealing with the physics of continuous media, especially when calculating forces and work.
The formula to find linear mass density r>is r>\(\lambda = \frac{m}{L}\), where \(m\) is the total mass, and \(L\) is the total length.
For our chain problem, the total mass is 30 kg, and the total length is 5 meters, resulting in a linear mass density of \(\lambda = 6 \text{ kg/m}\).
Each meter of the chain weighs 6 kg, which allows us to understand how the chain behaves when lifted.

Gravitational Force

Gravitational force is the attractive force exerted by the Earth on objects that have mass. When lifting a chain, understanding the effect of gravitational force on each segment of the chain is vital.
The force on a small piece of the chain at height \(h\) is influenced by both the chain's linear mass density and gravity.
The formula for the gravitational force on a small segment is \(F(h) = \lambda g (L - h)\), where:

• \(\lambda\) is the linear mass density,
• \(g\) is the gravitational acceleration (approximately 9.81 m/s² on Earth), and
• \(L - h\) gives the remaining length of the chain still on the ground.

Understanding gravitational force helps in analyzing how much work is needed to lift the chain.

Integral Calculation

Integral calculation is used in physics to find quantities like work, which involve adding up a continuous distribution of forces or effects.
When the chain is lifted, the work done involves calculating the force needed to lift each small piece, and integrating these forces over the distance lifted.
The work done \(W\) is found by integrating the force over the height lifted: \(W = \int_{0}^{2} F(h) \, dh\).
Substituting \(F(h)\) into the integral gives: \(W = 6(9.81) \int_{0}^{2} (5-h) \, dh\).
This integral represents the sum of the gravitational forces on each small segment of the chain over the height from 0 to 2 meters. After evaluation, the resulting work done is found to be 471.12 joules, which signifies the energy expended in lifting the chain.