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A train of 3000 KNHi all 1 ) A train of 3000 KN weight starts from rest and reached a velocity of 54 km/h after one min . if the frictional resistance of the track is 5N/KN of the train's weight . Find the tractive force developed by the engine ? I don't understand what is mean "tractive force " ?? [color=#FF0000]anyway I start solve the question w = 3000 kN mass = 305810 u = 0 v = 54 km/h v = u + at a = 0.25 F = m X a F = 305810 X 0.25 76452.5 N [/color] 2 ) A parachute weighing 500 N falling with uniform acceleration from rest, descends 5 m in the first 4s. Find the resultant of air on the parachute . |

Re: A train of 3000 KN1.) We need to find the final velocity in meters per second: Hence: Your answer is correct. 2.) First, we may find the net force on the parachute: Now, if we let be the resultant force of air on the parachute, we have: |

Re: A train of 3000 KNteacher said the answer for first one is not complete he said F = p - ( 5 X 3000 ) F = p - 15000 kN I don't know what he mean ? |

Re: A train of 3000 KNIt appears that he means you need to add the resistive force to the result to get the total. |

Re: A train of 3000 KNAre you mean tractive force = force + resistive force or what ? and I want know ,,, what is the difference between force which arr got and tractive force ? |

Re: A train of 3000 KNFor Q 2 ... Fnet = m Xa (500/9.81 ) X ( 0.625 ) = 31.85 N I don't know How you got ( 0.14 N ) |

Re: A train of 3000 KNHow did you get 0.625 for the acceleration? I get 1/360 using which comes from where . |

Re: A train of 3000 KNsee h = ut+1/2at^2 5 = 0 + 0.5a(4^)2 a = 5/8 which is 0.625 ?? where is the mistake ? |

Re: A train of 3000 KNOoops, I used the time from the first problem. :lol: Carry on... |

Re: A train of 3000 KNnow I know all the answer just this formula I don't know why you said F net = w - Fr can please explain to me fully |

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