# Hard mechanics questionWatch

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#1
An engine and a train weigh 250 tonnes. The engine exerts a pull of 1.5x10 to the 5 (sorry, can't write it in the format) Newtons. The resistance to motion is 1/100 of the weight. The train starts from rest and accelerated uniformly until it reaches a speed of 45km/h. At this point, brakes are applies until the train stops. Find the time take for the train to stop.

Please help - I've tried a couple of times but I can't see how the parts of the equation link (e.g I got the first acceleration as 0.49, and the deceleration as 0.59, but don't see how they link and am unsure if they're right - also, are you meant to use suvat about the train first accelerating?) Not too sure what to do.

Thank you
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1 year ago
#2
(Original post by jsjsjsjs333)
An engine and a train weigh 250 tonnes. The engine exerts a pull of 1.5x10 to the 5 (sorry, can't write it in the format) Newtons. The resistance to motion is 1/100 of the weight. The train starts from rest and accelerated uniformly until it reaches a speed of 45km/h. At this point, brakes are applies until the train stops. Find the time take for the train to stop.

Please help - I've tried a couple of times but I can't see how the parts of the equation link (e.g I got the first acceleration as 0.49, and the deceleration as 0.59, but don't see how they link and am unsure if they're right - also, are you meant to use suvat about the train first accelerating?) Not too sure what to do.

Thank you
0
#3
(Original post by Muttley79)
Yeah sure, give me a min
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#4
For the first part:

A driving force of 1.5x10 to the 5, being pulled backwards by the resistance to motion, so an overall force of:
1.5x10 to the 5 - 2500 = 147500 N

Then, I thought 147500= ma, and then a = 0.59.

I wasn't quite sure what to do next, because it looked like I should have used suvat, (e.g it tells you it started from rest and what it ended on, which was giving enough info for suvat). I didn't know the use in finding out the distance or time, so carried on and kinda just left this part behind...

Then, the car starts to brake. The braking force I worked out as 25,000, and the resistance to motion continues to act in the same direction, so:
147,500-25,000= 122500.

Again I used
122500 = ma
so a = 0.49

Then I used SUVAT,

and said
initial v = 12.5 (45km/h in m/s)
final v = 0
a = 0.49

and then used v=u+at to find t
0=12.5+0.49t
-12.5=0.49t
t=-25.5

But the time should't be negative and it's wrong anyways, so I'm not sure what to do. I knew the method was wrong because I just abandoned the first bit of info, but I can't figure out how they link.

Could you give me some tips?
(Original post by Muttley79)
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1 year ago
#5
(Original post by jsjsjsjs333)
For the first part:

A driving force of 1.5x10 to the 5, being pulled backwards by the resistance to motion, so an overall force of:
1.5x10 to the 5 - 2500 = 147500 N
Could you give me some tips?
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#6
(Original post by Muttley79)
26.2 seconds
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1 year ago
#7
(Original post by jsjsjsjs333)
26.2 seconds
It will be decelerating in the braking part of the motion - how did you get your value?
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#8
(Original post by Muttley79)
It will be decelerating in the braking part of the motion - how did you get your value?
I tried to find the total forces acting on it, so did the engine force - resistance to motion - braking force. Then said that the answer to that was equal to mass x acceleration, and divided by mass. Is this right?
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1 year ago
#9
(Original post by jsjsjsjs333)
I tried to find the total forces acting on it, so did the engine force - resistance to motion - braking force. Then said that the answer to that was equal to mass x acceleration, and divided by mass. Is this right?
There is no force driving the train forward - what is g in this question?
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#10
(Original post by Muttley79)
There is no force driving the train forward - what is g in this question?
Is there not a force from the engine? g is 9.8
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1 year ago
#11
(Original post by jsjsjsjs333)
Is there not a force from the engine? g is 9.8
It's braking ...
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#12
(Original post by Muttley79)
It's braking ...
Oh god of course
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#13
(Original post by Muttley79)
It's braking ...
How do you then find the force as it moves forward? As in when it starts braking, because they'll still be force forwards? Or do you not need this?
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1 year ago
#14
(Original post by jsjsjsjs333)
How do you then find the force as it moves forward? As in when it starts braking, because they'll still be force forwards? Or do you not need this?
The resistive force is 1/100 of the weight but the weight is given in tonnes not newtons.
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