Truck accelerating down a plane- A level Watch

Lola1244
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A truck is accelerating down a plane inclined at 30 degrees to the horizontal from rest. What is distance traveled after 4 seconds.
I used the suvat equations and got 78.48... but i took acceleration as 9.81.
Is this right because it is an inclined plane.
I tried to do trigonometry on the answer:
78.48/sin30 but that wasn't right answer???

THANK YOU FOR HELP I WILL REP BEST ANSWER
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Mr. Ee
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(Original post by Lola1244)
A truck is accelerating down a plane inclined at 30 degrees to the horizontal from rest. What is distance traveled after 4 seconds.
I used the suvat equations and got 78.48... but i took acceleration as 9.81.
Is this right because it is an inclined plane.
I tried to do trigonometry on the answer:
78.48/sin30 but that wasn't right answer???

THANK YOU FOR HELP I WILL REP BEST ANSWER
Instead of using acceleration as 9.81 try finding the component of the acceleration down the slope.

Using that you should be able to get the correct answer
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Smithenator5000
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(Original post by Lola1244)
A truck is accelerating down a plane inclined at 30 degrees to the horizontal from rest. What is distance traveled after 4 seconds.
I used the suvat equations and got 78.48... but i took acceleration as 9.81.
Is this right because it is an inclined plane.
I tried to do trigonometry on the answer:
78.48/sin30 but that wasn't right answer???

THANK YOU FOR HELP I WILL REP BEST ANSWER
Hello,

Acceleration down an inclined plane is equal to the acceleration due to gravity multiplied by the sin of the incline angle:

a = g sin(30o) = 1/2 g

You can now substitute all the known values into . . .

s = ut + 1/2 at2


. . . to get:

s = (0)(4) + 1/2 (1/2 g )(4)2 = 4g

If g is taken to be 9.81ms-2, s is 39.2m (3 significant figures).

I hope that this has been helpful.
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Lola1244
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(Original post by Smithenator5000)
Hello,

Acceleration down an inclined plane is equal to the acceleration due to gravity multiplied by the sin of the incline angle:

a = g sin(30o) = 1/2 g

You can now substitute all the known values into . . .

s = ut + 1/2 at2


. . . to get:

s = (0)(4) + 1/2 (1/2 g )(4)2 = 4g

If g is taken to be 9.81ms-2, s is 39.2m (3 significant figures).

I hope that this has been helpful.
Thank you so much for your help- I did eventually figure it out!
I was wondering if you could help me out with this problem:
a girl holding a weight stands on a set of weighing scales. When she moves up her arms rapidly to lift the weight up, the reading on the scale rises.

Is this because when she lifts her hands up rapidly she is accelerating the weight and thereby causing a force upwards so therefore due to her pushing the weight up, Newton's Third Law tells us that the weight will exert a force downwards on her???
So if the reading on the weighing scale decreases she has accelerated the weight downwards, therefore the weight has exerted a force on her upwards, reducing her weight. And if the reading on the weighing scale is constant then it means that she is not moving the weight as Newton's First Law tells us that if an object is stationary then the resultant force is 0.

Thank you for your help- the new a level questions are so hard :/
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Lola1244
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(Original post by Mr. Ee)
Instead of using acceleration as 9.81 try finding the component of the acceleration down the slope.

Using that you should be able to get the correct answer
Thanks I did do that
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Smithenator5000
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(Original post by Lola1244)
Thank you so much for your help- I did eventually figure it out!
I was wondering if you could help me out with this problem:
a girl holding a weight stands on a set of weighing scales. When she moves up her arms rapidly to lift the weight up, the reading on the scale rises.

Is this because when she lifts her hands up rapidly she is accelerating the weight and thereby causing a force upwards so therefore due to her pushing the weight up, Newton's Third Law tells us that the weight will exert a force downwards on her???
So if the reading on the weighing scale decreases she has accelerated the weight downwards, therefore the weight has exerted a force on her upwards, reducing her weight. And if the reading on the weighing scale is constant then it means that she is not moving the weight as Newton's First Law tells us that if an object is stationary then the resultant force is 0.

Thank you for your help- the new a level questions are so hard :/
Hello,

I believe that you have the right idea. When she is lifting the weight, she is exerting a force on it in the upwards direction. As you have already identified, this results in an equal and opposite force being applied to her due to Newton's third law. I believe that the opposite does indeed happen if she lets the weight fall, as the weight either stops contributing to the reading (if she merely releases it) or it subtracts from it (if she actively forces it downwards). If the weight is being held in a stationary position, the reading should not change, as the resultant force that is being exerted in the downward direction is constant; it is the weight of the girl plus the weight of the 'weight'. In theory, I believe that if the girl was constantly accelerating the weight upwards or downwards, the reading should not change during this time, as constant acceleration implies constant force. This of course would be hard to achieve. Apologies if any of this has not been clear. I would be happy to paraphrase it.

Anyway, I am glad that I have been helpful. May I ask, which syllabus do you use?
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Lola1244
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(Original post by Smithenator5000)
Hello,

I believe that you have the right idea. When she is lifting the weight, she is exerting a force on it in the upwards direction. As you have already identified, this results in an equal and opposite force being applied to her due to Newton's third law. I believe that the opposite does indeed happen if she lets the weight fall, as the weight either stops contributing to the reading (if she merely releases it) or it subtracts from it (if she actively forces it downwards). If the weight is being held in a stationary position, the reading should not change, as the resultant force that is being exerted in the downward direction is constant; it is the weight of the girl plus the weight of the 'weight'. In theory, I believe that if the girl was constantly accelerating the weight upwards or downwards, the reading should not change during this time, as constant acceleration implies constant force. This of course would be hard to achieve. Apologies if any of this has not been clear. I would be happy to paraphrase it.

Anyway, I am glad that I have been helpful. May I ask, which syllabus do you use?
Hello, thank you so much for your help- I have understood the majority of it However, I am a bit lost on the part where you said that if she was accelerating the weight upwards of downwards at a constant force the reading would not change? Surely a constant force upwards would create a constant force downwards that would add on to her weight and cause an increase on the reading? Could you please explain that?
I do understand that if she was moving the weight with constant velocity then reading would not change because Newton's First Law tells that on constant velocity resultant force is 0

I am doing the new AQA physics as level course- they have introduced new multiple choice questions which are rather tough :')
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Smithenator5000
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(Original post by Lola1244)
Hello, thank you so much for your help- I have understood the majority of it However, I am a bit lost on the part where you said that if she was accelerating the weight upwards of downwards at a constant force the reading would not change? Surely a constant force upwards would create a constant force downwards that would add on to her weight and cause an increase on the reading? Could you please explain that?
I do understand that if she was moving the weight with constant velocity then reading would not change because Newton's First Law tells that on constant velocity resultant force is 0

I am doing the new AQA physics as level course- they have introduced new multiple choice questions which are rather tough :')
Hello,

My sincere apologies; I must have been unclear. By 'would not change', I meant that between one point in time when that force is being applied and another when that same force is being applied, the reading would not change because the forces combine in exactly the same way at each point in time. I assume that you have interpreted my previous description as stating that the reading would not change with respect to the initial reading (with the weight at rest), however it of course would. Just for clarification, you are correct about the constant velocity resulting in a constant reading; it is the girl's weight plus the weight of the 'weight' because she must match the weight's weight in order for Newton's first law to remain applicable. I like to think about this as a special case for the constant acceleration (which has been re-explained earlier in this reply). Instead of having a positive or negative value, acceleration is 0. Therefore, the idea that constant force results in no change can be applied to it. It's not necessary to think of it in that way, however I find it helpful.

Once again, I am sorry if any of this is poorly explained; I have a list of excuses if you're interested. That being said, I still hope that I have managed to take the 'weight' off your shoulders . Is your syllabus the A or B syllabus? Don't worry- my friends don't particularly enjoy my curiosity either.
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Lola1244
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(Original post by Smithenator5000)
Hello,

My sincere apologies; I must have been unclear. By 'would not change', I meant that between one point in time when that force is being applied and another when that same force is being applied, the reading would not change because the forces combine in exactly the same way at each point in time. I assume that you have interpreted my previous description as stating that the reading would not change with respect to the initial reading (with the weight at rest), however it of course would. Just for clarification, you are correct about the constant velocity resulting in a constant reading; it is the girl's weight plus the weight of the 'weight' because she must match the weight's weight in order for Newton's first law to remain applicable. I like to think about this as a special case for the constant acceleration (which has been re-explained earlier in this reply). Instead of having a positive or negative value, acceleration is 0. Therefore, the idea that constant force results in no change can be applied to it. It's not necessary to think of it in that way, however I find it helpful.

Once again, I am sorry if any of this is poorly explained; I have a list of excuses if you're interested. That being said, I still hope that I have managed to take the 'weight' off your shoulders . Is your syllabus the A or B syllabus? Don't worry- my friends don't particularly enjoy my curiosity either.

Oh thank you so much! I get it now! Hahaha oh gosh physics jokes never get old ey Um I think the new AQA physics spec from 2015 only has one syllabus as far as I am aware! What board are you doing?
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Lola1244
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(Original post by Smithenator5000)
Hello,

My sincere apologies; I must have been unclear. By 'would not change', I meant that between one point in time when that force is being applied and another when that same force is being applied, the reading would not change because the forces combine in exactly the same way at each point in time. I assume that you have interpreted my previous description as stating that the reading would not change with respect to the initial reading (with the weight at rest), however it of course would. Just for clarification, you are correct about the constant velocity resulting in a constant reading; it is the girl's weight plus the weight of the 'weight' because she must match the weight's weight in order for Newton's first law to remain applicable. I like to think about this as a special case for the constant acceleration (which has been re-explained earlier in this reply). Instead of having a positive or negative value, acceleration is 0. Therefore, the idea that constant force results in no change can be applied to it. It's not necessary to think of it in that way, however I find it helpful.

Once again, I am sorry if any of this is poorly explained; I have a list of excuses if you're interested. That being said, I still hope that I have managed to take the 'weight' off your shoulders . Is your syllabus the A or B syllabus? Don't worry- my friends don't particularly enjoy my curiosity either.
I'm sorry to ask you again but I have a slight confusion- you know when she lifts the weight upwards and thus creates an upwards force resulting in an equal and opposite downwards force... why do the two forces not just cancel out and as a result the reading would not change
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Smithenator5000
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(Original post by Lola1244)
Oh thank you so much! I get it now! Hahaha oh gosh physics jokes never get old ey Um I think the new AQA physics spec from 2015 only has one syllabus as far as I am aware! What board are you doing?
Hello,

I'm not quite sure how to efficiently respond to two posts but I'll certainly try. The board only using one syllabus surprises me and I would imagine that it would be an inter-board change if that was the case. The AQA website seems to imply that they still use the two syllabus system. I may be wrong. As far a I'm aware, my college use the OCR A syllabus, however my perception of reality has just been challenged! As for the physics jokes, I have enough to write a book. All I need is a publisher . . .
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Smithenator5000
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(Original post by Lola1244)
I'm sorry to ask you again but I have a slight confusion- you know when she lifts the weight upwards and thus creates an upwards force resulting in an equal and opposite downwards force... why do the two forces not just cancel out and as a result the reading would not change
Hello,

That is a good question, which I hear being asked all the time. When the girl exerts a force on the weight, the weight applies the force to her. Vectors, in general, can only cancel if they were applied to the same object. To paraphrase, the weight experiences the upward force, whilst the girl experiences the downward force.

PS: apologies are not necessary. I have not got that much else to do in all honesty.
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Lola1244
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(Original post by Smithenator5000)
Hello,

That is a good question, which I hear being asked all the time. When the girl exerts a force on the weight, the weight applies the force to her. Vectors, in general, can only cancel if they were applied to the same object. To paraphrase, the weight experiences the upward force, whilst the girl experiences the downward force.
Ah okay! got it completely- thank you so much!!!! I understand it
Hahahaha I look forward to buying the book lol
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Smithenator5000
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(Original post by Lola1244)
Ah okay! got it completely- thank you so much!!!! I understand it
Hahahaha I look forward to buying the book lol
I'm glad you understand . Good luck with the rest of your A levels.
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