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A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.45 and μk = 0.38, respectively.

The acceleration of gravity is 9.8 m/s2 .
What is the frictional force acting on the 25 kg mass?
Answer in units of N
What is the largest angle which the incline can have so that the mass does not slide down the incline?
Answer in units of ◦
What is the acceleration of the block down the incline if the angle of the incline is 29◦ ?
Answer in units of m/s2

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4 answers

  1. 24.2o what is the sign after 24.2?

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  2. degrees probably

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  3. a. Wt. = m * g 25kg * 9.8N/kg = 245 N.

    F1 = Force parallel to the incline.
    F2 = Force perpendicular to the incline.

    Q1. F1 = 245*sinA = u*F2
    245*sinA = 0.45*245*cosA
    Divide both sides by 245*cosA:
    sinA/cosA = 0.45
    Replace SinA/cosA with tanA:
    tanA = 0.45
    A = 24.2o = Angle of the incline.

    Fs=u*245*cosA=0.45*245*cos24.2=100.6 N.
    = Force of static friction.

    Q2. The largest angle is the angle at which the force of static friction is
    equal to F1. The angle is 24.2o.

    Q3. F1 = 245*sin29 = 118.8 N.
    F2 = 245*cos29 = 214.3 N.

    Fk = u*F2 = 0.38 * 214.3 = 81.4 N. =
    Force of kinetic friction.

    a=Fn/m = (F1-Fk)/m=(118.8-81.4)/25 = 1.49 m/s^2.

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  4. this is sd afae[a

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