I will most definitely reward a best answer. Thanks for the help! A 5.0-kg block

Question

I will most definitely reward a best answer. Thanks for the help!

A 5.0-kg block and a 4.0-kg block are connected by a 0.6-kg rod. The links between the blocks and the rod are denoted by A and B. A force F is applied to the upper block. 2) In Fig. 4.3, the blocks and rod assembly moves downward at constant velocity. The applied force F is closest to: A)88N B)90N C)92N D)96N E)94N 3) In Fig. 4.3, the blocks and rod assembly moves upward at constant velocity. The force in link A is closest to: A)43N B)47N C)39N D)45N E)41N 4) In Fig. 4.3, the applied force F equals 150 N. The force in link B is closest to: A)58N B)54N C)62N D)56N E)60N 5) In Fig. 4.3, the force in Link B is 40 N. The acceleration of the blocks and rod assembly, including direction, is closest to: A) zero B) 2.4 m/s^2, downward C) 2.4 m/s^2, upward D) 1.2 m/s^2, upward E) 1.2 m/s^2, downward 6) In Fig. 4.3, the blocks and rod assembly accelerates upward at 2.0 m/s^2. The force in Link A exceeds the force in link B by an amount that is closest to: A)7N B)8N C)6N D)10N E)9N

Explanation / Answer

Answer 2

9.6*9.8 = 94 N option E

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