A 6.0-kg box slides down an inclined plane that slopes 39° to the horizontal.

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 A 6.0-kg box slides down an inclined plane that slopes 39° to the horizontal. The box accelerates at a rate of 3.1 m/s2. What is the coefficient of kinetic friction between the box and the surface of the inclined plane?
a  0.40
b  0.55
c  0.16
d  0.48
Apr 23rd, 2015

Select the coordinate system so that the x-axis is directed down the slope and the y-axis is directed perpendicular to the slope. There are three forces acting on the box: the weight W =(W_x, W_y), the normal reaction force (0, N), and the friction force (F, 0). Since Wx = mgsinθ; Wy = –mgcosθ, where θ = 39°, we may write the components of the resultant force (mgsinθ - F,  –mgcosθ + N) = (ma, 0) where the acceleration is directed down the slope. Write the equations mgsinθ - F = ma and  –mgcosθ + N = 0. The friction force equals F = μN, where μ is the coefficient of the kinetic friction. Thus, F = mgsinθ - ma = μN = μmgcosθ and 

μ = (mgsinθ - ma)/mgcosθ = (gsinθ - a)/gcosθ = (9.81sin39° - 3.1)/(9.81cos39°)  = 0.40.

Answer: a)

Apr 24th, 2015

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