How Much Work Is Required to Push a 10 kg Steel Block Across a Table?

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SUMMARY

The discussion focuses on calculating the work required to push a 10 kg steel block across a steel table at a steady speed of 1.0 m/s for 3 seconds, using the coefficient of kinetic friction of 0.60. The work done is calculated using the formula W = Fd, resulting in 176.4 Joules. Additionally, the power output during this process is determined using P = Fv, yielding a power output of 58.8 Watts. These calculations are based on the principles of physics involving friction and motion.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concept of friction, specifically kinetic friction
  • Knowledge of basic physics equations for work and power
  • Ability to perform calculations involving force, distance, and velocity
NEXT STEPS
  • Study the impact of different materials on the coefficient of friction
  • Learn about the relationship between work, energy, and power in physics
  • Explore advanced applications of Newton's laws in real-world scenarios
  • Investigate the effects of varying speeds on work and power calculations
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Students studying physics, educators teaching mechanics, and anyone interested in understanding the principles of work and power in physical systems.

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Homework Statement



a. How much work must you do to push a 10 kg block of steel across a steel table at a steady speed of 1.0 m/s for 3 s? The coefficient of kinetic friction for steel on steel is 0.60.

b. What is your power output while doing so?

Homework Equations



W = Fd (where d = distance)
F = (fric. coeff)N (where N=normal force)
Normal force in this case = weight
w = mg

P = Fv


The Attempt at a Solution



a) w = mg
= 10(-9.8)
= -98

N = 98

F = (fric. coeff)(N)
= (0.6)(98)
= 58.8

W = Fd d = 3m
= (58.8)(3)
= 176.4 J

b)
P=Fv
= (58.8)(1)
=58.8W
 
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