Calculating Work for Stacking Books on a Table

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SUMMARY

The discussion focuses on calculating the work required to stack six identical books, each 4.0 cm thick and weighing 0.80 kg, on a table. The initial calculation incorrectly assumes a uniform height for lifting all books. The correct approach involves recognizing that each book is lifted to a different height, requiring a total work of 4.7 J, as derived from the formula U = mgy, where 'y' varies for each book. The key takeaway is the importance of calculating the individual heights for each book when stacking.

PREREQUISITES
  • Understanding of gravitational potential energy (U = mgy)
  • Basic knowledge of physics concepts related to work and energy
  • Ability to perform calculations involving mass, height, and gravitational acceleration
  • Familiarity with unit conversions (e.g., converting cm to meters)
NEXT STEPS
  • Study the concept of gravitational potential energy in depth
  • Learn about work-energy principles in physics
  • Explore problems involving variable heights in stacking or lifting scenarios
  • Practice calculations involving multiple objects and their respective heights
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Students studying physics, educators teaching mechanics, and anyone interested in understanding work calculations in real-world scenarios.

laxboi33
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Homework Statement



Six identical books, 4.0 cm thick and each with a mass of .80 Kg, lie individually on a flat table. How much work would be needed to stack the books one on top of the other.



Homework Equations


U= mgy



The Attempt at a Solution



total mass = .8kg x 6 = 4.8
G= 9.8
y= .004 * 6 = .024

(4.8)(9.8)(.024) = 1.12 J

the answer is 4.7 J. What am I doing wrong?
 
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y is not simply 6 x brick thickness.

The bricks are each raised a different height, in order to make them into a single stack of 6 bricks.
 
Youre doing work against gravity right? You have to lift one book to put it on the other book, a certain height. When you add a third book, has the height you have to lift it changed?
 

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