##### 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  0.40b  0.55c  0.16d  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.

Apr 24th, 2015

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Apr 23rd, 2015
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Apr 23rd, 2015
May 28th, 2017
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