- #1
dorothy
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- Homework Statement
- Does anyone knows how to solve these questions? Thanks a lot!
- Relevant Equations
- v=u+at
s=ut+1/2at^2
v^2=u^2+2as
Last edited:
SUVAT Problem: Box moving on inclined plane
- Thread starter dorothy
- Start date
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- Tags
- Box Inclined Inclined plane Plane Suvat
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Homework Help Overview
The discussion revolves around a problem involving a box moving on an inclined plane, specifically focusing on the SUVAT equations. Participants are addressing various parts of the problem, particularly part (c), which asks how far up the plane the box travels and its velocity at that point.
Discussion Character
- Mixed
Approaches and Questions Raised
- Some participants express uncertainty about their answers to parts (a) and (b) and seek validation. Others question the interpretation of the problem, particularly regarding the concept of projection and the definitions of acceleration and velocity. There is also discussion about the implications of changing conventions for acceleration.
Discussion Status
Participants are actively engaging with each other's ideas, questioning assumptions, and exploring different interpretations of the problem. Some guidance has been offered regarding the understanding of acceleration as a vector and the implications of the problem's setup, but no consensus has been reached on the answers.
Contextual Notes
There are indications of confusion regarding the definitions of terms and the conventions used in the problem, particularly concerning the direction of acceleration and the meaning of the point of projection.
- #2
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You seem to already have solved everything except (c), for which you have not shown any effort. What are your own thoughts regarding (c)?
- #3
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In part c, how far up the plane does the box go? What is its velocity at this point?dorothy said:
- #4
dorothy
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Hi, I want to whether my answers on a & b are correct or not? For (c), I don’t really get the question, like what is the point of projection? How’s the deceleration and acceleration do in this case (c)? Thank you.Orodruin said:
- #5
dorothy
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Before that, I want to ask whether I understand question8 correctly? Is it true that there is a guy throwing(projecting) a box horizontally and the box finally land on the inclined plane(like what I have drawn)? Thank you.Chestermiller said:
- #6
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I don't agree with your answer to a). Acceleration is a vector, not a scalar.dorothy said:
This also affects the answer I would give to b) i) I).
Thaty said, it's not clear that the question setter acknowledges that acceleration is a vector.
- #7
dorothy
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Should (a) be -4.905?PeroK said:
- #8
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That's what I would put. With the units as you have in your answer.dorothy said:
- #9
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PS It does say velocity ##5 \ m/s## up the incline, which infers a convention that up the incline is positive. I'd prefer, however, that the question setter be more explicit about this. E.g. say a velocity of ##+5 \ m/s## up the incline. Then there is no doubt.
- #10
dorothy
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I see. What about (b)? Are all of them correct?PeroK said:
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I don't agree with b i) I).dorothy said:
- #12
BvU
Science Advisor
Homework Helper
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I expect the exercise composer means ##5## m/s along the incline, not ##5## m/s in a horizontal diertion.dorothy said:
##\ ##
- #13
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If we look at part c) it's clear that up is positive. Note, however, that the question setter gets confused with motion down the slope and changes the convention. The acceleration down the slope should be ##-2 \ ms^{-2}##.
This seems to be a common cause of confusion: that negative acceleration is reducing speed (deceleration) and positive acceleration is increasing speed. But, if you think about it, this is not correct. Once we have established that up is positive, then the acceleration of gravity is ##-9.8 \ ms^{-2}## throughout projectile motion. It doesn't change to ##+9.8 \ ms^{-2}## for the descent. And, in fact, the SUVAT equation would no longer work if you change your convention half way through.
In this case, the acceleration does change so you have to reset your SUVAT equations. But, I suggest there is no need to change convention and make down positive.
This seems to be a common cause of confusion: that negative acceleration is reducing speed (deceleration) and positive acceleration is increasing speed. But, if you think about it, this is not correct. Once we have established that up is positive, then the acceleration of gravity is ##-9.8 \ ms^{-2}## throughout projectile motion. It doesn't change to ##+9.8 \ ms^{-2}## for the descent. And, in fact, the SUVAT equation would no longer work if you change your convention half way through.
In this case, the acceleration does change so you have to reset your SUVAT equations. But, I suggest there is no need to change convention and make down positive.
- #14
dorothy
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PeroK said:
May I know why bi) is not correct? I don’t know how to do it.
- #15
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Sorry I meant b ii) I.dorothy said:
The acceleration has a greater magnitude in this case.
- #16
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So higher acceleration gives smaller distance traveled until it turns around ... as OP stated.PeroK said:
- #17
dorothy
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But if i assume the theta is 45° (which is greater than 30°, tilited more), then i put it into -9.81sin45°. I get the new acceleration=-6.9ms^-2. Next, I put 6.9 into the equation and get the new max distance =1.8 which is smaller than the original 2.55PeroK said:
- #18
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That right. It's like having more powerful brakes: you stop in a shorter distance.dorothy said:
- #19
dorothy
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So if the plane is tilited more, it becomes steeper, the distance traveled will also become smaller. Does it mean that I am actually correct on bii)1)?Orodruin said:
- #20
dorothy
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Does it mean that bii) I) is correct ?PeroK said:
- #21
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Yes, sorry. I thought b ii) I was asking about the acceleration. I see now it's asking about the distance travelled.dorothy said:
- #22
dorothy
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No worries :) For part c, can you explain to me what is the meaning of reaching the point of projection? I don’t know which point it is. I’m super confused nowPeroK said:
- #23
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It was my fault.dorothy said:
I can only imagine that it means back at the bottom of the incline!
- #24
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Probably the point where it was released with the original speed, but on the way down.dorothy said:
- #25
dorothy
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PeroK said:
Thank you for your help! :)