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http://upload.lsforum.net/users/public/o7986YF-05-51_newj86.jpgA load of bricks with mass m1 = 16.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2 = 29.0 kg is suspended from the other end of the rope, as shown in the figure. (Figure 1) The system... 顯示更多 http://upload.lsforum.net/users/public/o7986YF-05-51_newj86.jpg A load of bricks with mass m1 = 16.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2 = 29.0 kg is suspended from the other end of the rope, as shown in the figure. (Figure 1) The system is released from rest. What is the magnitude of the upward acceleration of the load of bricks? ? m/s^2 What is the tension in the rope while the load is moving? ?N
最佳解答:
Net force = (29g - 16g) N = 13g N = 130 N where g is the acceleration due to gravity, taken to be 10 m/s^2 Acceleration = net-force/mass = 130/(29+16) m/s^2 = 2.889 m/s^2 Consider the load of bricks, use: force = mass x acceleration T - 16g = 16 x 2.889 where T is the tension in the rope T = 206 N
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Newton's law發問:
http://upload.lsforum.net/users/public/o7986YF-05-51_newj86.jpgA load of bricks with mass m1 = 16.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2 = 29.0 kg is suspended from the other end of the rope, as shown in the figure. (Figure 1) The system... 顯示更多 http://upload.lsforum.net/users/public/o7986YF-05-51_newj86.jpg A load of bricks with mass m1 = 16.0 kg hangs from one end of a rope that passes over a small, frictionless pulley. A counterweight of mass m2 = 29.0 kg is suspended from the other end of the rope, as shown in the figure. (Figure 1) The system is released from rest. What is the magnitude of the upward acceleration of the load of bricks? ? m/s^2 What is the tension in the rope while the load is moving? ?N
最佳解答:
Net force = (29g - 16g) N = 13g N = 130 N where g is the acceleration due to gravity, taken to be 10 m/s^2 Acceleration = net-force/mass = 130/(29+16) m/s^2 = 2.889 m/s^2 Consider the load of bricks, use: force = mass x acceleration T - 16g = 16 x 2.889 where T is the tension in the rope T = 206 N
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