Energy loss in a pulley, but cleverer

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SUMMARY

The discussion revolves around calculating energy loss in a pulley system involving a block of mass 0.7 kg and a pulley with a moment of inertia of 0.004 kg*m². The block descends 0.83 m, reaching a speed of 1.611 m/s, leading to the calculation of energy dissipated due to friction. The derived energy loss, Q, is 8.15 J, although there is confusion regarding the sign of this value, with another participant suggesting a work done by friction of -3.25 J. The calculations utilize principles of energy conservation and rotational dynamics.

PREREQUISITES
  • Understanding of Newtonian mechanics, specifically energy conservation.
  • Familiarity with rotational dynamics and moment of inertia.
  • Knowledge of kinematic equations related to motion.
  • Basic algebra for solving equations involving energy terms.
NEXT STEPS
  • Study the principles of energy conservation in mechanical systems.
  • Learn about the effects of friction in pulley systems.
  • Explore the relationship between linear and angular velocity in rotating systems.
  • Investigate the calculation of work done by friction in various mechanical contexts.
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Physics students, mechanical engineers, and anyone interested in understanding energy dynamics in pulley systems and the effects of friction on energy loss.

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Homework Statement



A block of mass 7.00×10-1kg is suspended by a string which is wrapped so that it is at a radius of 5.80×10-2m from the center of a pulley. The moment of inertia of the pulley is 4.00×10-3kg*m2. There is friction as the pulley turns. The block starts from rest, and its speed after it has traveled downwards a distance of D=0.830m is 1.611m/s. Calculate the amount of energy dissipated up to that point.

given m, r, I, v, D, find Q (energy lost)

Homework Equations



mgh=U(x)
1/2Iw^2
I/2mv^2


The Attempt at a Solution



-mgh=1/2Iw^2+1/2mv^2+Q
solve for Q, get 8.15 J lost. I'm not really sure why I'm unable to get the right answer...unless 8.15 is supposed to be negative. Any thoughts? Thanks so much =)!
 
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Hmm, your method is correct. I got -3.25 as the work done by friction. Show a bit more in depth how exactly you made your calculation.
 

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