Drinking juice A glass has circular cross sections that taper (linearly) from a radius of 5 cm at the top of the glass to a radius of 4 cm at the bottom. The glass is 15 cm high and full of orange juice. How much work is required to drink all the juice through a straw if your mouth is 5 cm above the top of the glass? Assume the density of orange juice equals the density of water.
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Work
Problem 6.7.31b
Textbook Question
Winding a chain A 30-m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and the chain has a density of 5kg/m.
b. How much work is required to wind the chain onto the cylinder if a 50-kg block is attached to the end of the chain?
Verified step by step guidance1
Identify the physical setup: A 30-meter chain with linear density 5 kg/m hangs vertically, and a 50-kg block is attached at the bottom. The goal is to find the work required to wind the entire chain and the block onto the cylinder (i.e., lift them up).
Express the weight of the chain per unit length as \(\lambda = 5 \text{ kg/m}\), so the mass of a segment of length \(x\) meters is \(m(x) = 5x\) kg. The weight force of this segment is \(W(x) = 9.8 \times 5x = 49x\) Newtons, where 9.8 m/s² is the acceleration due to gravity.
Set up the integral for the work done to lift the chain: Consider a small segment of the chain at a distance \(x\) from the bottom (where \(x=0\) is the bottom and \(x=30\) is the top). This segment must be lifted a distance of \(x\) meters to be wound onto the cylinder. The infinitesimal work to lift this segment is \(dW = \text{force} \times \text{distance} = 49 \, dx \times x = 49x \, dx\).
Account for the 50-kg block at the bottom: The block has a weight of \$50 \times 9.8 = 490\( Newtons and must be lifted the full 30 meters. The work to lift the block is \)W_{block} = 490 \times 30$.
Write the total work as the sum of the work to lift the chain and the work to lift the block: \(W_{total} = \int_0^{30} 49x \, dx + 490 \times 30\). Evaluate the integral to find the total work required.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Work Done by a Variable Force
Work is the integral of force over a distance. When the force changes with position, as in lifting parts of a chain, we calculate work by integrating the weight of each segment over the distance it moves. This approach accounts for the varying force needed to lift different lengths of the chain.
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Weight and Density Relationship
The weight of an object is the product of its mass and gravitational acceleration. For a chain with uniform density, mass is density times length. Understanding this helps determine the force exerted by the chain at any point, which is essential for calculating the work done in lifting it.
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Superposition of Forces in Work Calculation
When multiple forces act in a system, such as the chain's weight and an attached block, the total work is the sum of work done against each force. This means calculating the work to lift the chain and the block separately, then adding them to find the total work required.
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