1. The problem statement, all variables and given/known data A block of mass m1 = 21.5 kg is at rest on a plane inclined at θ = 30.0° above the horizontal. The block is connected via a rope and massless pulley system to another block of mass m2 = 17.7 kg, as shown in the figure. The coefficients of static and kinetic friction between block 1 and the inclined plane are μs = 0.109 and μk = 0.086, respectively. If the blocks are released from rest, what is the displacement of block 2 in the vertical direction after 1.47 s? Use positive numbers for the upward direction and negative numbers for the downward direction. 2. Relevant equations D = v_i*t + (1/2)a*t^2 F = ma 3. The attempt at a solution For mass 2, Positive is going upwards. For mass 1, Positive is going down the ramp. (sum) 1. F_m1_x = [(m_1)g*sin(30)] - [(m_1)g*cos(30)] - T = (m_1)a (sum) 2. F_m2_y = T - (m_2*g) = (m_2)a --> T = (m_2*a) + (m_2*g) Plugging equation 2 into 1 for T, we get: [(m_1)*g*sin(30)] - [(m_1)g*cos(30)] - [(m_2*a) + (m_2*g)] = m_1*a ---> [(m_1)*g*sin(30)] - [(m_1)g*cos(30)] - (m_2*g) = m_1*a - [(m_2*a) ---> g(m_1*sin(30) - m_1*cos(30) - m_2) = a(m_1 - m_2) a = [g(m_1*sin(30) - m_1*cos(30) - m_2)] / (m_1 - m_2) plugging in the values, we get a = -66.01 m/s^2 cool, now we can find how far it travels in 1.47 s. D = v_i*t + (1/2)a*t^2 d = 0 + (1/2)(-66.01 m/s^2)*(2.1609) d = -71.32m That's not right. Unsure what I'm doing wrong :/ I also tried the block going up the ramp, but that didn't work either. This is my first post here, any help is appreciated, thanks!