What Are the Mechanical Advantages and Efficiencies in an Inclined Plane System?

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SUMMARY

The discussion focuses on calculating the mechanical advantages and efficiencies of an inclined plane system involving a 1-meter board inclined at 10 degrees. The ideal mechanical advantage (IMA) is calculated as 2450, while the actual mechanical advantage (AMA) is determined to be 2.45. The theoretical efficiency is found to be approximately 41.76%, and the experimental efficiency is calculated at 25.88%. Key calculations involve the forces acting on a block of mass 122.5g and a weight of 50g, with input and output work derived from these values.

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  • Understanding of mechanical advantage concepts
  • Knowledge of inclined plane physics
  • Familiarity with basic physics equations for work and force
  • Ability to perform calculations involving mass and gravitational force
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  • Study the principles of mechanical advantage in simple machines
  • Learn about the calculations for work and energy in physics
  • Explore the effects of friction on inclined planes
  • Investigate real-world applications of inclined planes in engineering
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Students studying physics, educators teaching mechanics, and engineers interested in the applications of inclined planes in design and efficiency analysis.

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Homework Statement



a 1meter board is inclined at 10 degrees with a height of .3 meters. a block with mass of 122.5g is placed at the bottom connected to a string which is connected to a pulley at the top. if a weight of 50g is hung from the pulley and the block moves with constant velocity, what is the ideal and actual mechanical advantage(theoretical), predicted efficiency(theoretical), input force(tension), input work, output work, AMA(experimental), and efficiency(experimental).


Homework Equations





The Attempt at a Solution

 
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Ideal mechanical advantage= output force/input force=122.5g/.05g=2450Actual mechanical advantage= net output force/input force=22.5g/50g=0.45Theoretical efficiency= output work/input work=122.5g*9.8m/s^2*sin10°*.3m/50g*9.8m/s^2=.4176Input force=MA*input force=2450*.05g=122.5gInput work=input force*distance=122.5g*9.8m/s^2*sin10°*.3m=2.7735JOutput work=output force*distance=50g*9.8m/s^2*sin10°*.3m=.7177JAMA(experimental)= output force/input force=122.5g/50g=2.45Efficiency(experimental)= output work/input work=.7177J/2.7735J=.2588
 

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