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1.A block accelerates at 3.1 m/s2 down a plane inclined at an angle 24.0◦. Find μk between the block and the in- clined plane. The acceleration of gravity is 9.81 m/s2 . 2.A 1200 kg car moves along a horizontal road at speed v0 = 20.1 m/s. The road is wet, so the static friction coefficient between the tires and the road is only μs = 0.189 and the kinetic friction coefficient is even lower, μk = 0.1323. The acceleration of gravity is 9.8 m/s2 . What is the shortest possible stopping dis- tance for the car under such conditions? Use g = 9.8 m/s2 and neglect the reaction time of the driver. Answer in units of m. Thank you for the help!

Verified Answer :

  1.  Fp = Mgsin24 = 0.407Mg. 
    Fn = Mg
    Cos24 = 0.914Mg. 
    Fk = u0.914Mg. 

    Fp-Fk = M
    a. 
    0.407Mg-u0.914Mg = Ma.
    Divide by Mg: 
    0.407-0.914u = a/g = 3.1/-9.81, 
    0.407-.914u = -0.316, uk = 0.791.

    2.  Mg = 12009.8 = 11,760 N. = Wt. of car. = Normal force(Fn). 
    Fp = Mgsin 0 = 0. 
    Fk = uk
    Fn = 0.132311,760 = 1556 N. 

    Fp-Fk = M
    a. 
    0-1556 = 1200a, a = -1.30 m/s^2. 

    V^2 = Vo^2 + 2a*d. 
    0 = (20.1)^2 - 19.6d