Two masses are being pulled up a 30.0 degrees incline by a force F parallel to the incline.The acceleration down the incline is 1.00 m/s² and the velocity is down the incline.The force is applied to a 200 kg mass and a string connects the 200 kg mass to a 150 kg mass.The coefficient of kinetic friction is 0.200.The force F is A) 771 N. B) 875 N. C) 1,050 N. D) 1,960 N. E) 2,330 N.

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Quizplus · Accepted Answer

The force of friction acting on the 200 kg mass is:$$f_k=\mu_k n=\mu_k(200	ext{ kg})(9.8	ext{ m/s}^2)\cos30\degree=392	ext{ N}$$The force of friction acting on the 150 kg mass is:$$f_k=\mu_k n=\mu_k(150	ext{ kg})(9.8	ext{ m/s}^2)\cos30\degree=294	ext{ N}$$The net force acting on the 200 kg mass is:$$F-f_k-200	ext{ kg}(9.8	ext{ m/s}^2)\sin30\degree=200	ext{ kg}(1	ext{ m/s}^2)$$$$F=392	ext{ N}+1960	ext{ N}=2352	ext{ N}$$The net force acting on the 150 kg mass is:$$T-f_k-150	ext{ kg}(9.8	ext{ m/s}^2)\sin30\degree=150	ext{ kg}(1	ext{ m/s}^2)$$$$T=294	ext{ N}+1470	ext{ N}=1764	ext{ N}$$Therefore, the force F is 2352 N.

Two Masses Are Being Pulled Up a 30