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 r-soy June 9th, 2012 11:30 AM

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 .

 MarkFL June 9th, 2012 12:44 PM

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}$

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}$

 r-soy June 10th, 2012 06:31 AM

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 ?

 MarkFL June 10th, 2012 08:21 AM

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.

 r-soy June 10th, 2012 08:37 AM

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 ?

 r-soy June 10th, 2012 11:21 AM

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 )

 MarkFL June 10th, 2012 11:30 AM

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 $v_0=0$.

 r-soy June 10th, 2012 11:40 AM

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 ?

 MarkFL June 10th, 2012 11:58 AM

Re: A train of 3000 KN

Ooops, I used the time from the first problem. :lol:

Carry on...

 r-soy June 10th, 2012 12:05 PM

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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