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A 5.00-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.

ADraw a free-body diagram for each block and identify the action-reaction forces between the blocks.
BDetermine the tension in the string and the magnitude of the acceleration of the 10-kg block

A 5.00-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 class=

Verified Answer :

All the forces are correct except that the counter-force in the 10.0 kg block must be 2um1g + um2g. The reaction force from the friction of the 5.00 kg block is um1g, and the friction force from the floor is u(m1 + m2)g.

The 5.00 kg block must be static. T1 = um1g = 0.20059.80 = 9.80 N.The net force of the 10.0 kg block is 45.0 - 20.2059.80 - 0.20109.80 = 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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