The two blocks shown start from rest and it is observed that the velocity at blo
ID: 2263183 • Letter: T
Question
The two blocks shown start from rest and it is observed that the velocity at block A is 5 ft/s after it has moved through 6 ft. Assuming the pulleys are massless and frictionless, determine the coefficient of kinetic friction between block A and the incline.Explanation / Answer
acceleration of block A is twice that of B. => aA = 2* aB let acceleration of block B be 'a' and that of block A be '(2*a)' vA = 5 ft/s = 5*0.3048 m/s = 1.524 m/s v0 = 0 s = 6 ft = 6*0.3048 m = 1.8288 m v^2 - v0^2 = 2*a*s => 1.524^2 - 0 = 2*(2*a)*1.8288 => a = 0.3175 m/s^2 (upwards) weight of block A = 200*0.453*9.8 = 887.88 N weight of block B = 300*0.453*9.8 = 1331.82 N for block A; T - mg*sin(theta) - u*mg*cos(theta) = mA *aA => T - 887.88*sin30 - u*887.88*cos30 = 200*0.453*(2*a) => T - 443.94 - 768.93*u = 90.6* 2*0.3175 => T = 768.93*u + 501.47 for block B; mg - T' = ma => 1331.82 - T' = 300*0.453*a => 1331.82 - T' = 135.9*0.3175 => T' = 1288.67 also, since the pulley is massless. for the pulley, T' = 2*T so, 2*T = 1288.67 => T = 644.34 T = 768.93*u + 501.47 => 644.34 = 768.93*u + 501.47 => u = 0.186
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