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Phy-motion(2)
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(a) Resistance force = 30 x 48 N = 1440 N Force develped by the engine = 1440 N Power of engine = 1440 x 48 watts = 69120 watts (b) Resistance force = 30 x 40 N = 1200 N Net force acting on the car by the engine = (1440 - 1200) N = 240 N Hence, acceleration = 240/1200 m/s2 = 0.2 m/s2 (c) Let v be the speed force exerted by the engine = 69120/v N hence, 69120/v + 1200g.sin(3) = 30v solve for v gives v = 60 m/s
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When a car,of mass 1200kg, travels at a speed of v ms^-1m it experiences a resistanceforce of magnitude 30v newtons.The car hasa max constant speed of 48 ms^-1 on a straight horizontal road.a) Showthat the max power of the car is 69120 Wb) The car is travelling along a straight horizontal road.Findthe max possible... 顯示更多 When a car,of mass 1200kg, travels at a speed of v ms^-1m it experiences a resistanceforce of magnitude 30v newtons.The car hasa max constant speed of 48 ms^-1 on a straight horizontal road.a) Showthat the max power of the car is 69120 Wb) The car is travelling along a straight horizontal road.Findthe max possible acceleration of the car when it is travelling at a speed of 40ms^-1c) Thecar starts to descend a hill on a straight road which is inclined at an angleof 3° to the horizontal. Find the max possible constant speed of the car as ittravels on this road down the hill.最佳解答:
(a) Resistance force = 30 x 48 N = 1440 N Force develped by the engine = 1440 N Power of engine = 1440 x 48 watts = 69120 watts (b) Resistance force = 30 x 40 N = 1200 N Net force acting on the car by the engine = (1440 - 1200) N = 240 N Hence, acceleration = 240/1200 m/s2 = 0.2 m/s2 (c) Let v be the speed force exerted by the engine = 69120/v N hence, 69120/v + 1200g.sin(3) = 30v solve for v gives v = 60 m/s
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