Recent content by dazedconfused

By using Newton's 2nd law, d=0.5at2 (since v0 is zero), I got an a=0.0556 m/s^2 Now, since they're on a pulley, a= (m2g-m1g) / (m2+m1) Plugging in, I get: 0.0556 = [(100kg*9.8m/s^2)-(m1*9.8m/s^2)] / 100kg + m1 rearranging, 5.56 + 0.0556 m1 = 980 - 9.8 m1...

Homework Statement The 100 kg block shown in the diagram takes 6.0s to reach the floor 1.0m below after being released from rest. What is the mass of the block on the left?Homework Equations Newton's Third Law F=ma F (a on b) = -F (b on a)The Attempt at a Solution I don't even know where to...