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

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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). 
  m/s2

Explanation / Answer

Net force on crate is given by Fnet = Fapplied - Ffriction

Since crate is pushed with a constant speed, so acceleration produced is zero. a=0

Also Fnet = ma = 0

So Fapplied = Ffriction

(a) F cos20o = µkN

     F cos20o = µk (mg + F sin20o)

    270 cos20o = µk (1480 + 270 sin20o)

   µk = 0.161

(b) Fnet = Fapplied - Ffriction

    ma = F cos20o - µk N

    (W/g)a = F cos20o - µk (mg - F sin20o)

    (1480/9.8) a = 270 cos20o - 0.161 (1480 - 270 sin20o)

    a = 0.2 m/s2

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