A 100-kg box is placed on a ramp.As one end of the ramp is raised,the box begins to move downward just as the angle of inclination reaches 25°.What is the coefficient of static friction between box and ramp?
A) 0.15
B) 0.27
C) 0.47
D) 0.95

Question

Quizplus · Accepted Answer

The force due to gravity acting on the box is given by Fg = mg, where m is the mass of the box and g is the acceleration due to gravity. 
Fg = (100 kg)(9.81 m/s^2) = 981 N 
The force acting parallel to the ramp is given by Fpara = Fg sinθ, where θ is the angle of inclination of the ramp. 
Fpara = (981 N)sin(25°) = 416 N 
The force of static friction acting opposite to the direction of motion is given by Fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force acting perpendicular to the surface. 
N = Fg cosθ = (981 N)cos(25°) = 889 N 
Therefore, 
Fs ≤ μsN 
Fs = Fpara = 416 N 
Substituting for Fs, 
μsN = 416 N 
μs(889 N) = 416 N 
μs = 0.468
The closest answer choice to this is C (0.47), so the correct answer is C.

A 100-Kg Box Is Placed on a Ramp