My Math Forum A train of 3000 KN

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 June 9th, 2012, 11:30 AM #1 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 A train of 3000 KN Hi 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 .
 June 9th, 2012, 12:44 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: A train of 3000 KN 1.) $F=ma=m$$\frac{v_f-v_i}{t}$$=\frac{w}{g}$$\frac{v_f-v_i}{t}$$$ We need to find the final velocity in meters per second: $v_f=54\text{ \frac{km}{hr}}\cdot\frac{1000\text{ m}}{1\text{ km}}\cdot\frac{1\text{ hr}}{3600\text{ s}}=15\text{ \frac{m}{s}}$ Hence: $F=\frac{3000\text{ kN}}{9.81\text{ \frac{m}{s^2}}}$$\frac{(15-0)\text{ \frac{m}{s}}}{60\text{ s}}$$\approx76.45\text{ kN}$ Your answer is correct. 2.) First, we may find the net force on the parachute: $F_{\text{net}}=ma=m$$\frac{2x}{t^2}$$=\frac{w}{g}\ (\frac{2x}{t^2}\)=\frac{500\text{ N}}{9.81\text{ \frac{m}{s^2}}}$$\frac{2\cdot5\text{ m}}{\(60\text{ s}$$^2}\)=\frac{1250}{8829}\:\text{N}$ Now, if we let $F_r$ be the resultant force of air on the parachute, we have: $F_{\text{net}}=w-F_r$ $F_r=w-F_{\text{net}}=$$500-\frac{1250}{8829}$$\text{ N}=\frac{4413250}{8829}\:\text{ N}\approx499.86\text{ N}$
 June 10th, 2012, 06:31 AM #3 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A train of 3000 KN teacher 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 ?
 June 10th, 2012, 08:21 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: A train of 3000 KN It appears that he means you need to add the resistive force to the result to get the total.
 June 10th, 2012, 08:37 AM #5 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A train of 3000 KN Are 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 ?
 June 10th, 2012, 11:21 AM #6 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A train of 3000 KN For Q 2 ... Fnet = m Xa (500/9.81 ) X ( 0.625 ) = 31.85 N I don't know How you got ( 0.14 N )
 June 10th, 2012, 11:30 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: A train of 3000 KN How did you get 0.625 for the acceleration? I get 1/360 using $a=\frac{2x}{t^2}$ which comes from $x=\frac{1}{2}at^2+v_0t$ where $v_0=0$.
 June 10th, 2012, 11:40 AM #8 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A train of 3000 KN see h = ut+1/2at^2 5 = 0 + 0.5a(4^)2 a = 5/8 which is 0.625 ?? where is the mistake ?
 June 10th, 2012, 11:58 AM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: A train of 3000 KN Ooops, I used the time from the first problem. Carry on...
 June 10th, 2012, 12:05 PM #10 Senior Member   Joined: Oct 2009 Posts: 895 Thanks: 1 Re: A train of 3000 KN now 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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