Question

A m, - 5.124, block is placed on top of a m=10.0-49 block (figure). A horizontal force of 450n is applied to the 10.0-kg block, and the 5.12-kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0.189. (a) Draw a free body diagram for each block and identify the action-reaction forces between the blocks. (Submit a file with a maximum siz of 1 MB) Choose File No file chosen This answer has not been graded yet. (b) Determine the tension in the string. (Enter the magnitude only.) N Determine the magnitude of the acceleration of the 10.0-kg block. m/s2 Need Help?

Answer 1

m, = 5.12kg M2 = 10.0kg u= 0.189 → F= 45.ON M2 (M. +M2)g fi= UN = umia = 0.1&9 x 5-12 x10= 9.67 N u(m, +M2) = f2 = u N₂ = 0.189 X 15.12*10= 28.57N Me HF fi Mig (M, TM2) g upper block is at rest so T=fi T= UN, = umig = 9.67N [T=9.670 equation of motion for mass M₂ Tafi-f2 = M₂ a 45-9.67-28.57= loa 6.76=100 (a= 0.676 m/s2