A5.00-kg block is placed on top of a 10.0-kg block (fig. p5.44). a horizontal force of 45.0 n is applied to the 10-kg block, and the 5-kg block is tied to the wall. the coefficient of kinetic friction between all moving surfaces is 0.200.
(a) draw a free-body diagram for each block and identify the action-reaction forces between the blocks.
(b) determine the tension in the string and the magnitude of the acceleration of the 10-kg block

All the forces are correct except that the counter-force in the 10.0 kg block must be 2*u*m1*g + u*m2*g. The reaction force from the friction of the 5.00 kg block is u*m1*g, and the friction force from the floor is u*(m1 + m2)*g.

The 5.00 kg block must be static. T1 = u*m1*g = 0.200*5*9.80 = 9.80 N.The net force of the 10.0 kg block is 45.0 - 2*0.20*5*9.80 - 0.20*10*9.80 = 5.80 N. Thus, the acceleration of the 10.0 kg block is 5.80 / 10 = 0.580 m/s^2.

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Centrapedial force is the answer