How Is Work Calculated When Pushing a Desk Across a Rough Surface?

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SUMMARY

In the discussion, Amadeus calculates the work done while pushing a desk across a rough surface with a mass of 94.0 kg. The coefficients of friction are 0.800 for static friction and 0.620 for kinetic friction. The correct approach involves using the coefficient of kinetic friction after the desk starts moving, leading to an acceleration of 7.84 m/s². The final calculation for work done is corrected to account for the kinetic friction, resulting in a total work of 66,000 Joules over 6 seconds.

PREREQUISITES
  • Understanding of Newton's Second Law (f=ma)
  • Knowledge of friction coefficients (static and kinetic)
  • Ability to calculate work (w=d*f)
  • Familiarity with kinematic equations (D=1/2at²)
NEXT STEPS
  • Study the effects of different friction coefficients on work calculations
  • Learn how to apply Newton's laws in real-world scenarios
  • Explore kinematic equations for varying acceleration scenarios
  • Investigate the relationship between force, mass, and acceleration in physics
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Students in physics, educators teaching mechanics, and anyone interested in understanding the principles of work and friction in physical systems.

terichristine
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Homework Statement


Amadeus pushes horizontally on a desk that rests on a rough wooden floor. The coefficient of static friction between the desk and the floor is 0.800 and the coefficient of kinetic friction is 0.620. The desk has a mass of 94.0 kg. He pushes just hard enough to barely get the desk moving and continues pushing with that force for 6.00 s. What work does he do on the desk during that time (HINT: Since once movement starts, friction changes to sliding friction, which is less than maximum static friction, the desk will be accelerating. SO, use the 2nd law to find the acceleration, then find out far it went in 6 s...)?


Homework Equations



u*mg
f=ma
w=d*f
D=1/2at^2

The Attempt at a Solution



I used:
u*mg=.8*94*9.8=736.96N
f=ma=736.96/94=acceleration=7.84
and displacement=1/2at^2=(.5)*(7.84)*(6^2)=141.2
w=d*f=141.2*736.96=104059 and that's wrong. I don't know what I am doing wrong. Thanks.
 
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terichristine said:

Homework Statement


Amadeus pushes horizontally on a desk that rests on a rough wooden floor. The coefficient of static friction between the desk and the floor is 0.800 and the coefficient of kinetic friction is 0.620. The desk has a mass of 94.0 kg. He pushes just hard enough to barely get the desk moving and continues pushing with that force for 6.00 s. What work does he do on the desk during that time (HINT: Since once movement starts, friction changes to sliding friction, which is less than maximum static friction, the desk will be accelerating. SO, use the 2nd law to find the acceleration, then find out far it went in 6 s...)?


Homework Equations



u*mg
f=ma
w=d*f
D=1/2at^2

The Attempt at a Solution



I used:
u*mg=.8*94*9.8=736.96N
f=ma=736.96/94=acceleration=7.84
and displacement=1/2at^2=(.5)*(7.84)*(6^2)=141.2
w=d*f=141.2*736.96=104059 and that's wrong. I don't know what I am doing wrong. Thanks.

Try using mu k, coefficient of kinetic friction, because the only work done is while the force displacing the box is acting. It said this in the hint. So 0.62 instead of 0.80
 

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