# Forces and Newton's Laws - A level Maths Question

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#1
Trying to solve these problems for a while now and they are confusing me. Anyone have any idea on how to solve these?

1) An object is hung from a spring balance suspended from the roof of a lift. When the lift is descending with uniform acceleration 0.8ms^-2 the balance indicates a weight of 245N. When the lift is ascending with uniform acceleration ams^-2 the reading is 294N. Find the value of a.

2) An engine and train weigh 250 tonnes. The engine exerts a pull of 1.5 x 10^5N. The resistance to motion is 1/100 of the weight of the train and its braking force is 1/10 of the weight. The train starts from rest and accelerates uniformly until it reaches a speed of 45kmh^-1. At this point the brakes are applied until the train stops. Find the time taken for the train to stop to the nearest second.

Thank you
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1 month ago
#2
(Original post by Quinton dj)
Hi, I can help with question 2
Let's start by making a summary like this:

They have giving us Mass of train (not the weight)= 250ton= 250000kg= 2.5x10^5 kg
The driving forward force =1.5x10^5N= 150000N
The resistance of motion is 1/100 of the [weight of the train]
braking force is 1/10 of the [weight of train]
highest speed= 45km/h = 12.5m/s
Now, the weight of train W=ma = 250000x10=2.5x10^6N (for simplicity, I'll use powers of ten. That way easy to follow and do on calculator)

So, braking force= 2.5x10^6x 1/10= 2.5x10^5N
Resistance force= 2.5x10^6x 1/100= 2.5x10^4N

Now we need acceleration between rest to final speed to find the time between when the train started accelerating and when it reach the highest speed of 45km/h.

So, we have a forward force driving the train forward and a resistance force decreasing the speed. We know Force=mass x acceleration F=ma
Therefore we can say forward Force minus resistance Force=mass x acceleration

fF - rF=ma 1.5x10^5 - 2.5x10^4=2.5x10^5a solve to find a= 125000/250000 =0.5m/s^2

So use formula of acceleration: a=v-u/t. Since u is zero(initial velocity) a= final velocity / time
therefore time = final velocity / acceleration = 12.5/0.5= 25seconds

For the time between braking, velocity starts with 12.5m/s and decreases to 0m/s so u=12.5 and v=0
But this time, we have a braking force plus the resistance force. So what we will do is add them together=
2.5x10^5+2.5x10^4= 2.75x10^5N backwards force.

Now, we have no forward force from the engine but a resistance force of friction and braking. So fF - rF=ma fF=0
So, 0-rF=ma our equation= -rF=ma let's add the numbers;
-2.75x10^5=2.5x10^5a solve a= -2.75x10^5/2.5x10^5=-1.1m/s^2 (the negative indicates deceleration which is happening right now as there is no forward force)

Now find time using a=v-u /t. Remember, our final velocity= 0 so a= -u /t rearange to make t the subject and you should get t= -u/a
Now add the numbers into equation t = -12.5/-1.1= 11.36.....repeatedly (see how important it is to get your signs right)

Finally, add the times together = 11.36+ 25.00=36.36seconds (round it off to 3significant figures) so time = 36.4seconds
And that is that, hope this can still help
This was posted four years ago and please note we do not post solutions - it's against forum rules
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