SUVAT Problem: Box moving on inclined plane

AI Thread Summary
The discussion revolves around solving a physics problem involving a box moving on an inclined plane, specifically focusing on part (c) regarding the distance the box travels up the incline and its velocity at that point. Participants express confusion about the acceleration and the conventions used in the problem, particularly whether acceleration should be treated as a vector. Clarifications are provided about the implications of different angles on acceleration and distance traveled, with an emphasis on maintaining consistent conventions throughout the calculations. The conversation highlights the importance of understanding the point of projection and how it relates to the box's motion. Overall, the thread emphasizes the need for clarity in problem statements and the correct application of SUVAT equations.
dorothy
Messages
39
Reaction score
1
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
5942A852-3C0E-4DC7-8E2D-68DEF1620E96.jpeg
 
Last edited:
Physics news on Phys.org
You seem to already have solved everything except (c), for which you have not shown any effort. What are your own thoughts regarding (c)?
 
dorothy said:
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

View attachment 299486
In part c, how far up the plane does the box go? What is its velocity at this point?
 
Orodruin said:
You seem to already have solved everything except (c), for which you have not shown any effort. What are your own thoughts regarding (c)?
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.
 
Chestermiller said:
In part c, how far up the plane does the box go? What is its velocity at this point?
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.
B622F996-33C4-4380-B9B7-95E7A0686790.jpeg
 
dorothy said:
Hi, I want to whether my answers on a & b are correct or not?
I don't agree with your answer to a). Acceleration is a vector, not a scalar.

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.
 
  • Like
Likes dorothy
PeroK said:
I don't agree with your answer to a). Acceleration is a vector, not a scalar.

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.
Should (a) be -4.905?
 
dorothy said:
Should (a) be -4.905?
That's what I would put. With the units as you have in your answer.
 
  • Like
Likes dorothy
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
PeroK said:
That's what I would put. With the units as you have in your answer.
I see. What about (b)? Are all of them correct?
 
  • #11
dorothy said:
I see. What about (b)? Are all of them correct?
I don't agree with b i) I).
 
  • #12
dorothy said:
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.
View attachment 299489
I expect the exercise composer means ##5## m/s along the incline, not ##5## m/s in a horizontal diertion.

:welcome:

##\ ##
 
  • Like
Likes Chestermiller and PeroK
  • #13
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.
 
  • #14
PeroK said:
I don't agree with b i) I).

May I know why bi) is not correct? I don’t know how to do it.
 
  • #15
dorothy said:
May I know why bi) is not correct? I don’t know how to do it.
Sorry I meant b ii) I.

The acceleration has a greater magnitude in this case.
 
  • #16
PeroK said:
Sorry I meant b ii) I.

The acceleration has a greater magnitude in this case.
So higher acceleration gives smaller distance traveled until it turns around ... as OP stated.
 
  • #17
PeroK said:
Sorry I meant b ii) I.

The acceleration has a greater magnitude in this case.
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 2.55
 
  • #18
dorothy said:
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 2.55
That right. It's like having more powerful brakes: you stop in a shorter distance.
 
  • #19
Orodruin said:
So higher acceleration gives smaller distance traveled until it turns around ... as OP stated.
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)?
 
  • #20
PeroK said:
That right. It's like having more powerful brakes: you stop in a shorter distance.
Does it mean that bii) I) is correct ?
 
  • #21
dorothy said:
Does it mean that bii) I) is correct ?
Yes, sorry. I thought b ii) I was asking about the acceleration. I see now it's asking about the distance travelled.
 
  • #22
PeroK said:
Yes, sorry. I thought b ii) I was asking about the acceleration. I see now it's asking about the distance travelled.
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 now
 
  • #23
dorothy said:
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.
It was my fault.

I can only imagine that it means back at the bottom of the incline!
 
  • #24
dorothy said:
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 now
Probably the point where it was released with the original speed, but on the way down.
 
  • #25
PeroK said:
It was my fault.

I can only imagine that it means back at the bottom of the incline!

Thank you for your help! :)
 
Back
Top