8.4.54/9.09 points | Previous AnswersSerCP10 4.P.041.My Notes Question PartPoint

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8.4.54/9.09 points |  Previous AnswersSerCP10 4.P.041.My Notes

Question PartPointsSubmissions Used

A 1,480-N crate is being pushed across a level floor at a constant speed by a force 

F

 of 270 N at an angle of 20.0°below the horizontal, as shown in the figure a below.

(a) What is the coefficient of kinetic friction between the crate and the floor? (Enter your answer to at least three decimal places.) 
  

(b) If the 270-N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in the figure b, what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part a

for part b I got 0.2 and it's not accepting the answer, so I posted it here and they got 0.2 as well. does some get a different anser

Explanation / Answer

f = Mu * N

Mu = 270cos 20/ (1480 + 270sin20)

Mu = 0.161

Acceleration of the crate is Mu*g = 9.81*0.161 = 1.58 m/s^2

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