Mass 2 m is attached to mass m with a light cable, strung over a pulley of mass
ID: 1567166 • Letter: M
Question
Mass 2m is attached to mass m with a light cable, strung over a pulley of mass m, and momnet of inertia I = mr^2 . Mass 2m rests on a wedge of inclination , compressing a spring with constant k to -x past the springs equilibrium position x0. The system starts at rest, but is not in equilibrium. Assuming -x and k are large enough to move 2m up the incline, what must the coefficient of kinetic friction uk between mass 2m and the incline be so that mass 2m travels distance d past x0 before stopping? What must the coefficient of static friction us be so that after 2m travels d , it stays at rest? Answer in terms of m, r, , k, d, x, and x0.
Explanation / Answer
Normal force = mg * cos(thetha)
Friction force = uk * mg * cos(thetha)
Net energy change in system = Friction losses
(1/2) * k * (x2 - d2) - (2m)*g*(x+d)
+ mg(x+h) = uk * mg * cos(thetha) * (x+d)
K*d + 2mgsin(thetha) - mg < us * 2mg * cos(thetha)