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標題:
Physics(mechanics)
發問:
An elevator and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42 m.
最佳解答:
Use equation of motion: v^2 = u^2 + 2as with v = 0 m/s, u = 12 m/s, s = 42 m, a = ? hence, 0 = 12^2 + 2a(42) a = -1.714 m/s^2 (the -ve sign indicates a retardation) Let T be the cable tnesion, use net-force = mass x acceleration 1600g - T = 1600 x 1.714 T = 1600 x (g -1.714) N = 13,257 N Alternatively, you may use "energy approach" Use: work done = change of kinetic energy (1600g - T) x 42 = (1/2).(1600).(12^2) solve for T gives T = 13,257 N
Physics(mechanics)
發問:
An elevator and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the elevator, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42 m.
最佳解答:
Use equation of motion: v^2 = u^2 + 2as with v = 0 m/s, u = 12 m/s, s = 42 m, a = ? hence, 0 = 12^2 + 2a(42) a = -1.714 m/s^2 (the -ve sign indicates a retardation) Let T be the cable tnesion, use net-force = mass x acceleration 1600g - T = 1600 x 1.714 T = 1600 x (g -1.714) N = 13,257 N Alternatively, you may use "energy approach" Use: work done = change of kinetic energy (1600g - T) x 42 = (1/2).(1600).(12^2) solve for T gives T = 13,257 N
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