Physics

A cardboard box of unknown mass is sliding upon a mythical frictionless surface.
The box has a velocity of 4.56 m/s when it encounters a bit of friction. After sliding 0.700m, the box has a velocity of 3.33 m/s.
What is the coefficient of friction of the surface?

Am I right to think the work energy theorem needs to be used? The friction reduces the velocity which means it must have removed kinetic energy. However, I don't know how to find the the coefficient of friction as I don't know the mass. Obviously, my thinking is off somewhere along the line but I'm not sure where exactly.

Any help is greatly appreciated.

Thank you in advance.

asked by Grace
  1. M*g = Wt. of box = Normal force(Fn).
    Fp = Mg*sin o = 0. = Force parallel to the surface.

    Fk = u*Fn = u*M*g.

    V^2 = Vo^2 + 2a*d = (3.3)^2.
    (4.56)^2 + 2a*0.7 = 10.9, a = -7.07 m/s^2.

    Fp-Fk = M*a.
    0-u*Mg = M*(-7.07),
    0-u*M*9.8 = -7.07M,
    Divide both sides by -9.8M:
    u = 0.722.


    posted by Henry

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