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.
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.
1 answer
1 answer