Determine the coefficient of friction

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SUMMARY

The discussion focuses on determining the coefficient of friction for a 10kg cardboard box sliding across a level floor, covering a distance of 6.0m and stopping in 2.2 seconds. The relevant equations include kinematic equations and Newton's second law, specifically \[\Delta d={{v}_{0}}t+\frac{1}{2}a{{t}^{2}}\] and \[{{F}_{net}}=ma\]. The solution reveals that the acceleration is calculated using the relationship \[ma=-\mu mg\], leading to a coefficient of friction of 0.26 when the box's initial velocity is considered. The discussion emphasizes the importance of correctly identifying initial conditions in motion problems.

PREREQUISITES
  • Understanding of kinematic equations, specifically \[\Delta d={{v}_{0}}t+\frac{1}{2}a{{t}^{2}}\]
  • Familiarity with Newton's second law, \[{{F}_{net}}=ma\]
  • Knowledge of the concept of friction and coefficient of friction
  • Basic algebra skills for solving equations with multiple variables
NEXT STEPS
  • Study the derivation of the coefficient of friction in various scenarios
  • Learn about kinematic equations in-depth, focusing on their applications in real-world problems
  • Explore the effects of different surfaces on the coefficient of friction
  • Investigate how to experimentally determine the coefficient of friction using different materials
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone interested in understanding motion and friction in practical applications.

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Homework Statement


A small 10kg cardboard box is thrown across a level floor. It slides a distance of 6.0m, stopping in 2.2s. Determine the coefficient of friction between the box and the floor.


Homework Equations


\[\Delta d={{v}_{0}}t+\frac{1}{2}a{{t}^{2}}\]
\[{{F}_{net}}=ma\]


The Attempt at a Solution


\[ma=-\mu mg\]
\[10a=-\mu 10(9.8)\]

\[6={{v}_{0}}(2.2)-\frac{1}{2}a{{(2.2)}^{2}}\]

How would I go about finding the acceleration?
The answer given is 0.26, which means the initially velocity of the cardboard box is zero. Is it okay to assume that the box is initially going at zero velocity and then decelerates to zero velocity due to friction?
 
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Write down the equation for the velocity of the box. What can you conclude from it? Since the box is thrown, obviously the initial velocity can't be zero.
 
One more equation to use is vf = vi+at, thus,
-vi = at
since vf = 0.
Now there are two equation with two unknowns (when considering the equations you have posed). You may now solve for acceleration.
 

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