Question

show all the work please

2. A 3.80-kg hanging block is pulling against a 2.50-kg block on a tabletop, as shown, via a string over a pulley that is massless and frictionless. At the same time, a constant external force F = 5.50 N is pulling the tabletop block to the right, as shown. There is a static friction coefficient is = 0.75, and kinetic friction coefficient dx = 0.15, with the tabletop (a) The blocks are initially stationary. With a calculation, determine whether the blocks stay stationary (b) If the blocks stay stationary, find the tension in the string; OR If the blocks start moving, find the acceleration of the blocks. I

Answer 1

M2 Let mi T 3.80 kg M2 = 2.50kg F = 5.50 N Ms = 0.7 Mk 0.15 T M₂9 mig @) Since initially both blocks are stationary = mig = 3.80*9.80 N 37:24 m2 = 2.50 Kg For block of mass f+Fx = + Ms M₂ Ms Mag 5.50 +0.75 x 2.50X9.80 N 23.875 N will move ft - - F+ Fx <T, So blocks will move

a mig g mass 2150 kg - (6) Let tension in the String is T and accelaration of the blocks is for block of mass m, = 3.80 kg mo T ma (2) for block of m2 = T-F- MKM 2 g Mz a ... (ii) Let's add c) and Yui's mi g- & - MK M2g = (M + M2) a (M, - MK M2) g- mit m2 (3.80 Q 0.18° X2°50) *9180- 5.50] (3.80 + 2050) 4845 m/s2 a = F 1 m/s2 n NOW from (0) eq we have T= mi (g-a) 3.80 ( 9180– 4:45 N 20:33 N