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A small solid was shoved up a plane which is inclined at an angle of . Find the coefficient of friction of the plane if the time of descent was twice the time of ascent.
Details
- Assume that slipping does not take place.
The answer is 0.161.
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1 solution
Using Newton's Second Law:
Ascent: ma1 = mg sin(15) + u mg cos(15) => a1 = g [sin(15) + u cos(15)]; Descent: ma2 = mg sin(15) - u mg cos(15) => a2= g [sin(15) - u cos(15)];
If the inclined distance is x and the solid starts from rest in both cases, then:
Ascent: x = (1/2) * a1 * t^2; Descent: x = (1/2) * a2 * (2t)^2;
and coupling these two kinematics equations for distance:
2x/t^2 = a1 = 4*a2 (i);
or g [sin(15) + u cos(15)] = 4 * g [sin(15) - u cos(15)];
Solving for the coefficient of friction, u, above yields:
u = (3/5) * [sin(15) / cos(15)] = (3/5)*tan(15) = 3/[5 * cot(15)] = 3/(5 * 3.732) = 0.161