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Types of force acting at once- show mass is 440kg

I’ve been gliding through this section as it seems more relatable and more common sense but this question has tripped me up a little.

See question below:
8CD82365-4485-48FC-A857-21ECED001374.jpeg


As f=ma, if force =1200-320=880 this is consistent throughout.

At point A it passes at speed of 12.

It accelerates to 26 in 7 seconds. I believe I need to find a(acceleration) that proves mass=440kg..

If I said 880=440a then this tells me acceleration is 2. I need to find this value I believe before it reaches point a.

Obviously, I’m a little confused, so sorry if my wording doesn’t make sense

FCE6AF72-E354-4D50-BAFB-2FE4941DBF54.jpeg

Would I be correct in saying the intial velocity is 12 when t=0?
(edited 1 year ago)

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Original post by KingRich
I’ve been gliding through this section as it seems more relatable and more common sense but this question has tripped me up a little.

See question below:
8CD82365-4485-48FC-A857-21ECED001374.jpeg


As f=ma, if force =1200-320=880 this is consistent throughout.

At point A it passes at speed of 12.

It accelerates to 26 in 7 seconds. I believe I need to find a(acceleration) that proves mass=440kg..

If I said 880=440a then this tells me acceleration is 2. I need to find this value I believe before it reaches point a.

Obviously, I’m a little confused, so sorry if my wording doesn’t make sense

FCE6AF72-E354-4D50-BAFB-2FE4941DBF54.jpeg

Would I be correct in saying the intial velocity is 12 when t=0?

Youve pretty much done it v = u+at and f=ma so
26 = 12 + 7a
880 = ma
(edited 1 year ago)
Reply 2
Original post by mqb2766
Youve pretty much done it v = u+at and f=ma so
26 = 12 + 7a
880 = ma


Ahh, you know what. I had a silly moment. I legit mis-read 26ms^-1 as the acceleration value and not as final velocity :rolleyes:. I’m sure if I’d come had a little break and re-read I would have realised but I’ve got deadlines
Reply 3
Original post by mqb2766
Youve pretty much done it v = u+at and f=ma so
26 = 12 + 7a
880 = ma


Without sounding completely dumb.

Part b, would suggest if the driving force is being reduced, then there’s a braking force involved.

Part a showed that acceleration was 2ms^-2 to reach 26ms^-1….

Hence f=ma would be 1200-f(new force)=(440)(-2) so, that our new force =320N
Original post by KingRich
Without sounding completely dumb.

Part b, would suggest if the driving force is being reduced, then there’s a braking force involved.

Part a showed that acceleration was 2ms^-2 to reach 26ms^-1….

Hence f=ma would be 1200-f(new force)=(440)(-2) so, that our new force =320N

Not quite. Right answer, but wrong reasoning.
If its travelling with constant velocity, what is the acceleration? So how is the driving force related to the resistance? Note its a "state" question, so no working is involved.

There is no braking here, just less driving force.
(edited 1 year ago)
Reply 5
Original post by mqb2766
Not quite. Right answer, but wrong reasoning.
If its travelling with constant velocity, what is the acceleration? So how is the driving force related to the resistance? Note its a "state" question, so no working is involved.

There is no braking here, just less driving force.

As per Newton’s first law, an object remains at constant velocity or at rest unless acted on by a force.

Hence, zero acceleration or deceleration. I’ll get back to this. Food
Original post by KingRich
As per Newton’s first law, an object remains at constant velocity or at rest unless acted on by a force.

Hence, zero acceleration or deceleration. I’ll get back to this. Food

Correct, but you should be able to simply write the answer down as the resistance is 320 N so the driving force must be ... for the resultant force to be zero (zero acceleration).
Reply 7
Original post by mqb2766
Correct, but you should be able to simply write the answer down as the resistance is 320 N so the driving force must be ... for the resultant force to be zero (zero acceleration).


You see I’ve been questioning that due to the wording of the question. Obviously when you take away the driving force you’re left with the resistance force, I.e the resistance of the road on your tires when you’re driving.

However, as the law states no force is acting on the object when the velocity is constant, then why does the resistance force still remain?

Unless the law specifically means a force that is caused by someone/something like an action/an active force and ignores any a force that’s natural. Ah, do you understand what I’m trying to say? Lol
Original post by KingRich
You see I’ve been questioning that due to the wording of the question. Obviously when you take away the driving force you’re left with the resistance force, I.e the resistance of the road on your tires when you’re driving.

However, as the law states no force is acting on the object when the velocity is constant, then why does the resistance force still remain?

Unless the law specifically means a force that is caused by someone/something like an action/an active force and ignores any a force that’s natural. Ah, do you understand what I’m trying to say? Lol


In newton 2, you have
f = ma
where f is the resultant (or total) force acting on the body. Here
f = driving - resistance
as driving is in the direction of motion and resistance is in the opposite direction. So if the resultant force is 0, then there is zero acceleration.
Original post by KingRich
….as the law states no force is acting on the object when the velocity is constant…

The law states no ‘resultant force’ acts on the object when velocity is constant. There can be forces, but they just have to cancel out.
Reply 10
Original post by mqb2766
In newton 2, you have
f = ma
where f is the resultant (or total) force acting on the body. Here
f = driving - resistance
as driving is in the direction of motion and resistance is in the opposite direction. So if the resultant force is 0, then there is zero acceleration.

Right, right. It makes sense now.
Original post by KingRich
Right, right. It makes sense now.


Its the same scenario as you reaching terminal velocity when falling from a plane when gravity and air resistance balance so there is zero acceleration.
Reply 12
Original post by mqb2766
Its the same scenario as you reaching terminal velocity when falling from a plane when gravity and air resistance balance so there is zero acceleration.


Is this what they call equilibrium? I believe that’s the next part of the section.


just to finish the question lol part c.

I’m correct In approaching this as a multi-stage question? Hence, I should find the displacement for part a first or?
Original post by KingRich
Is this what they call equilibrium? I believe that’s the next part of the section.


just to finish the question lol part c.

I’m correct In approaching this as a multi-stage question? Hence, I should find the displacement for part a first or?

Yes, you need to include part a in the displacement calculation, though you could recognize that the veloicty-time graph is a trapezium (assuming you model the deceleration due to resistance correctly) which is the same another question you asked about so you could get the displacement from the trapezium in 1 calculation.
Reply 14
Original post by mqb2766
Yes, you need to include part a in the displacement calculation, though you could recognize that the veloicty-time graph is a trapezium (assuming you model the deceleration due to resistance correctly) which is the same another question you asked about so you could get the displacement from the trapezium in 1 calculation.


D6BF7FC2-8434-4982-AC0E-6108AFEBAEB3.jpeg

this is the graph that I found where point a starts at t=0.

Does this look about right to you?

Oh, it drives for 12 seconds so 12=19 on the graph
(edited 1 year ago)
Original post by KingRich
D6BF7FC2-8434-4982-AC0E-6108AFEBAEB3.jpeg

this is the graph that I found where point a starts at t=0.

Does this look about right to you?

Yes, but I forgot that the "trapezium" doesnt start at zero velocity so it is a bit different. So do
first trapezium + rectangle + triangle
areas to get the total distance/displacement covered. You can form them as suvat
s = (u+v)t/2
for the first trapezium
s = ut
for the rectangle
s = (u+0)t/2
for the triangle. Thinking of it as suvat is probably overkill in this case and finding areas is porbably simpler to visualize? Note you don't need to know the acceleration/deceleration if you use
s = average velocity * time
"suvat" or area.
(edited 1 year ago)
Reply 16
Original post by mqb2766
Yes, but I forgot that the "trapezium" doesnt start at zero velocity so it is a bit different. So do
first trapezium + rectangle + triangle
areas to get the total distance/displacement covered. You can form them as suvat
s = (u+v)t/2
for the first trapezium
s = ut
for the rectangle
s = (u+0)t/2
for the triangle. Thinking of it as suvat is probably overkill in this case and finding areas is porbably simpler to visualize? Note you don't need to know the acceleration/deceleration if you use
s = average velocity * time
"suvat" or area.


Mmm, that’s what I’ve done. So treated it as a 3 stage question.

So, first stage: s=133m second stage: s=312
third stage:

is my deceleration correct, as I seem to be coming up short in the distance. I found that base was 25.9 seconds but this suggests that the final stage is 338m

The book states 910m total i seem to be off by at least 300m


Edit:

So, I originally calculated the deceleration based on t=12 second but that would suggest that the second stage stops at v=0.

The third stage only gives an initial velocity and an end velocity. It doesn’t tell you how many t later that it stops. I have to somehow find out the hypotenuse with only a given height.

Or, I consider the natural resistance is 320 with zero driving force then as f=ma, theoretically speaking 320=440a would tell me resistance =0.7272
(edited 1 year ago)
Reply 17
Original post by mqb2766
Yes, but I forgot that the "trapezium" doesnt start at zero velocity so it is a bit different. So do
first trapezium + rectangle + triangle
areas to get the total distance/displacement covered. You can form them as suvat
s = (u+v)t/2
for the first trapezium
s = ut
for the rectangle
s = (u+0)t/2
for the triangle. Thinking of it as suvat is probably overkill in this case and finding areas is porbably simpler to visualize? Note you don't need to know the acceleration/deceleration if you use
s = average velocity * time
"suvat" or area.


So, I reproached the question using my brain this time and I believe I found my way.

5CF48924-8BEC-48E3-B0D4-8FF6BB5DACEF.jpeg
Not checked the numbers carefully, but looks about right. In the last part a should be negative as its decelerating so youd have
0 = u^2 - 2s*0.72727
which is why you have a negative sign kicking about which gets "forgotten" about.
Reply 19
Original post by mqb2766
Not checked the numbers carefully, but looks about right. In the last part a should be negative as its decelerating so youd have
0 = u^2 - 2s*0.72727
which is why you have a negative sign kicking about which gets "forgotten" about.


I knew the displacement would be positive, so I guess I kinda just ignored that fact but I recall for the future. A little proud moment for myself 😅

I definitely feel like talking out loud or discussing helps me find my way.
(edited 1 year ago)

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